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THE  LIBRARY 

OF 

THE  UNIVERSITY 

OF  CALIFORNIA 


GIFT  OF 

Kenneth  K.  Noble 


HOLWES  &00K  CO, 

-^3  8,  Main  St. 

Los  Angeles 


Erkata 


C'ORKIGE 


11 

1     with 

with- 

13 

6     that  s 

tliat  is 

16 

17-26     base  etc. 

angle   A  by  drawing  the  line  AD. 
The  triangle  .456' will  thus  be  di- 
vided into  two  triangles  ABD  and 
ACD,  which  will  be  equal  (No.  11); 
and  therefore  the  angles  B  and  C 
opposite  the  common  side  J.i>  will 
be  equal.  Q.  E.  D. 

Cor.    The  line  that  bisects  the 
angle  at  the  vertex  of  an  isosceles 
trianirle,  bisects  the  base  of  the  tri- 
angle, and  is  perpendicular  to  it. 

20 

24    2(n— 2),  or  2;i— 4. 

2(r<— 2)  right  angles. 

32 

21     POHPOSITION 

PROPOSITION 

33 

25     DCB 

DOE 

27    DCB 

DCE 

(54 

14,-15,-16   and  an  equal 

included  between  equal  angles,  arc 

angle  at  ^,  are  equal  (No.27,Cor'l; 

equal  (No.  12); 

20,-21  and  an  equal  angle 

betv/een  two  equal  angles, 

at^, 

68 

19     CD 

AD 

87 

15     EF 

EA 

94 

19    ao  :  AO 

Vo:  VO 

.. 

28    a^'.AO' 

W'.VG' 

103 

18  and  20    ? 

0 

h 
6 

104 

^    1 

h 
6 

124 

26  (No.  124, 

(No.  134, 

V 


Digitized  by  tine  Internet  Archive 

in  2008  with  funding  from 

IVIicrosoft  Corporation 


http://www.archive.org/details/elementsofgeometOObaymrich 


ELEMENTS 


GEOMETEY 


FOR  THE  USE  OF  BEGINNERS. 


BT 

JOSEPH  BAYMA,  S.J., 

Prof  essor  of  Mathematics  in  Santa  Clara  CoUege,  S.  J.,  Santa  Clara,  California. 


SAN  FBANCISCO  : 

A.    WALDTEUFEL, 

737  MARKET  STREET. 

1885. 


Copyright,  1885, 
By  a.    WALDTEUFEL. 


GIFT 


CfM£3 


CONTENTS. 


Preface,     ...••••.•••         5 

Preliminary  notions,         ..••••••         7 

BOOK  I. 
Angles  and  triangles,        ..•..•«•         9 

BOOK  II. 
Rectilinear  areas,  and  relations  of  similar  figures,    ...       28 

BOOK   III. 
The  circle  and  the  regular  polygons,        .       .       .       •       »       43 

BOOK    IV. 
Geometric  constructions, ,       ,       69 

BOOK  V. 

Planes,  and  polyhedral  angles,  .       •       •       •       t        •       71 

8 

709 


4  CONTENTS. 

PAGE 

BOOK   VI. 
Surfaces  and  volumes  of  polyhedrons, 86 

BOOK   VII. 
Surfaces  and  volumes  of  round  bodies, 105 

BOOK  VIII. 
Spherical  geometry,    .........      120 


PKEFAOE. 


The  present  treatise,  like  the  preceding  one  on  Al- 
gebra, has  been  written  for  the  instruction  of  the 
younger  students  of  Santa  Clara  College.  Among  the 
many  text-books  of  Geometry  published  in  this  coun- 
try which  we  have  read  or  used  in  past  years,  some 
have  been  found  too  diffuse,  others  full  of  cumber- 
some matter,  others  again  too  comprehensive  and  too 
difficult  for  beginners.  Comprehensive  books  are  very 
useful  in  the  hands  of  those  whose  minds  are  already 
formed ;  but  experience  has  taught  us  that  a  judicious 
parsimony  proves  more  successful  in  encouraging  the 
mental  efforts  of  young  beginners,  amid  the  many  dif- 
ficulties arising  from  the  giddiness  natural  to  their 
age,  as  well  as  from  the  number  of  their  scholastic 
duties. 

It  has  been  my  purpose  to  offer  to  this  class  of 
students  an  elementary  Geometry  sufficiently  devel- 
oped to  embrace  all  the  theorems  that  have  a  real 
importance,  and  yet  sufficiently  concise  and  plain  to 
make  it  possible  to  learn  it  all  within  six  or  seven 
months  in  the  first  year  of  the  mathematical  curri- 
culum.    The  treatise,  of   course,  contains  nothing  that 

6 


6  PREFAGJS. 


is  really  new ;  it  is  not,  however,  a  mere  compilation. 
The  order  and  method  which  I  have  followed,  and  the 
proofs  which  I  have  chosen,  are  such  as  will,  in  my 
opinion,  materially  lessen  the  labor,  and  at  the  same 
time  foster  the  industry  of  the  young,  in  the  acquisi- 
tion of  mathematical  knowledge.  Such,  at  least,  is  the 
goal  I  have  striven  to  reach. 

J.  B. 


ELEMENTS  OF  GEOMETRY. 


PEELIMINAKY  NOTIONS. 

1,  Geometry  is  that  branch  of  Mathematics  which 
treats  of  lines,  surfaces,  solids,  angles,  and  their  manifold 
relations. 

A  line  is  a  length  in  space  having  no  breadth  and  no 
thickness. 

A  surface  is  an  extension  in  length  and  breadth,  but 
having  no  thickness.  A  surface  when  measured  by  a 
given  unit  is  called  an  area. 

A  solid  is  a  magnitude  having  length,  breadth,  and 
thickness.  Length,  breadth,  and  thickness  are  called  the 
three  dimensions  of  geometric  bodies.  A  solid  when 
measured  by  a  given  unit  is  called  a  volume. 

An  angle  is  the  divergence,  or  difference  of  direction, 
of  two  lines,  with  reference  to  a  common  point  of  inter- 
section. 

2,  A  line  is  nothing  but  the  track  of  2i.  point  moving 
in  space.  A  point  has  no  dimensions;  but  a  moving 
point,  by  its  movement,  traces  a  line,  whose  length  mea- 
sures the  length  of  the  movement.  In  fact,  no  line  can 
be  drawn  but  by  moving  from  a  point  in  space  towards 
some  other  point  in  space  in  a  continuous  manner.  Lines 
are  either  straight  or  curved. 

A  straight  lins  is  a  line  whose  direction  is  one  and  the 
same  throughout. 


8  ELEMENTS  OF 


A  curved  line,  or  a  curve,  is  a  line  whose  direction  is 
continually  changing  by  imperceptible  degrees. 

Surfaces  are  either  jplane  or  curved,  A  plane  surface 
is  one  on  which  straight  lines  can  be  drawn  in  all  direc- 
tions. A  ciirved  surface  is  one  on  which  straight  lines 
cannot  thus  be  drawn. 

Angles  are  either  right  or  oblique  ;  and  oblique  angles 
are  either  acute  or  obtuse.  When  one  straight  line  meets 
another,  it  makes  with  it  two  angles,  which  are  called 
adjacent  angles.  If  these  two  angles  are  equal,  they  are 
called  right  angles;  but  if  unequal,  then  the  lesser  is 
called  an  acute,  and  the  greater  an  obtuse  angle. .  The 
lines  which  form  the  sides  of  an  oblique  angle  are  called 
oblique,  whilst  those  which  meet  at  right  angles  are  said 
to  hQ  perjpendicular  to  each  other. 

The  point  where  two  lines  meet  is  called  the  vertex 
of  the  angle  formed  by  them ;  and  the  angle  itself  may 
be  designated  by  a  letter  attached  to  the  vertex.  When, 
however,  two  or  more  angles  are  formed  about  the  same 
point,  then,  to  avoid  confusion,  every  angle  is  designated 
by  three  letters,  of  which  the  middle  one  marks  the  ver- 
tex of  the  angle,  and  the  other  two  the  directions  of  the 
sides. 


GEOMETRY. 


BOOK   I. 


ANGLES  AND    TRIANGLES. 

3.  Geometric  quantities  are  quantities  having  dimen- 
sions in  space.  All  sucli  quantities  are  continuous,  be- 
cause the  lines  which  represent  their  dimensions  {lengthy 
hreadth^  and  thickness)  are  the  traces  of  continuous  move- 
ments. Lines  may  be  compared  with  one  another  with 
regard  both  to  their  length,  and  to  their  relative  position. 
Their  position  implies  not  only  their  distances,  but  also 
their  mutual  inclination,  or  the  angle  contained  between 
them.  It  is  by  this  simplest  of  all  relations  between 
straight  lines  that  we  must  start  our  geometrical  inves- 
tigations. 


PEOPOSITIOI^  I. 

4:*  If  a  straight  line  meets  another^  the  sum  of  the 
adjacent  angles  will  he  equal  to  two  right  angles. 

Proof.  Let  DC  meet  AB 
at  C.  Drawing  CE  perpen- 
dicular to  AB,  we  have  these 
two  equations, 

DCA=ACE-\-ECD, 
BGD^ECB-ECD, 

which  being  added  together  give 

DCA-\-BCD  =  A  CE-\-  ECB. 

But  the  sum  ACE ^ ECB  is  equal  to  two  right  angles. 


^M- 


10  ELEMENTS  OF 


Therefore  the  sum  DCA-\-BCD  is  equal  to  two  right 
angles.     Q.  E.  D. 

Cor.  I.  If  one  of  the  adjacent  angles  is  right,  the  other 
also  is  right. 

Cor.  II.  The  sum  of  all  the  angles  formed  about  the 
same  point  on  the  same  side  of  a  line,  is  equal  to  two 
right  angles. 


PEOPOSITIOIT  11. 

5.  If  two  straight  lines  intersect  each  other,  the  oppo- 
site, or  vertical,  angles  will  he  equal. 

Proof.  Let^^andi>^in-  a, 
tersect  at  C.  The  angles  A  CD 
and  ECB  are  called  opposite 
or  vertical y  and  so  also  are  the 
angles  A CE and  DCB.  ^w 
(Ko.  4), 

A  CD  +  A  CE  =  two  right  angles, 
ECB  +  ACE=  two  right  angles  ; 

Therefore  ACD  +  A  CE=  ECB  +  A  CE;   and,  bj  re- 
duction, 

ACD  =  ECB.     Q.E.  D. 

It  may  be  shown,  in  a  similar  manner,  that  A  CE=DCB. 
Cor.  The  sum  of  all  the  angles  that  can  be  formed 
about  a  point  is  equal  to  four  right  angles. 


PKOPOSITIOIN^  III. 

^.  If  a  straight  line  intersects  two  pa/rallel  lines,  the 
sum  of  the  two  interior  angles  on  the  same  side  of  the 
intersecting  line  is  equal  to  two  right  angles. 

Proof.  Two  lines  are  said  to  be  pa/rallel  when  they 


aEOMETRY.  11 


lie  in  the  same  plane  in  a  similar  direction,  that  is,  with 
ont  any  inclination  towards  each  other.     Hence  parallel 
lines  keep  the  same  relative  distance  throughout  their 
extent,  and  can  never  meet,  how  far  soever  produced. 

Let,  then,  AB  and  CD 
be  parallel,  and  let  FE  in- 
tersect AB  at  G,  and  CD 
at  H,  Since  GB  and  HD 
run  in  the  same  direction, 
they  are  equally  inclined 
to  the  line  FE.  In  other 
words,  they  make  with  it 
equal  angles.     Consequently 

BGF^DHG, 

or,  adding  the  angle  BGIl  to  both  members  of  this 
equation, 

BGF^  BGH^  DHG  -^BGR. 

But  (:N'o.  4)  the  sum  BGF+BGR  is  equal  to  two 

right  angles.     Therefore  the  sum  2>^6^-|"^^^^s  ^^^^ 
equal  to  two  right  angles.     Q.  E.  D. 

Cor.  The  alternate  angles  between  parallel  lines  are 
equal.  Alternate  angles  are  those  which  lie  at  the  oppo- 
site sides  of  the  intersecting  line,  the  one  having  for  its 
sides  the  intersecting  line  and  one  of  the  two  parallel 
lines,  the  other  having  for  its  sides  the  intersecting  line 
and  the  other  of  the  two  parallel  lines.  Such  are  the  an- 
gles A  Gil  and  GIID,  both  interior,  and  such  are  also 
the  angles  FGB  and  CHE,  both  exterior.  It  is  evident 
t\\2X  AGII=GIID ;  for  both  these  angles  are  equal  to 
the  same  angle  FGB, 


12  ELEMENTS  OF 


PEOPOSITION  ly. 

7.  Two  lines  will  he  pa/rallel  if  the  sum  of  the  two 
interior  angles  on  the  same  side  of  an  intersecting  line  ,  JiJ. 

he  equal  to  two  right  angles,  -"Jl '  ^ 

Proof  Let  the  lines  J.^  P        /  ^; 

and  CD  be  intersected  by  ^  c,  r  :- 

a  line  FE  so  that  the  sum    y^__ Qx/fl^         d 

of  the  angles  BGII  and 
GHD  shall  be  equal  to  two  q_ 
right  angles.  Then,  since 
(Ko.  4)  the  sum  of  the  ad- 
jacent angles  BGH  and 
BGF\^  also  equal  to  two  right  angles,  we  shall  have 

BGII\-  GHD^zBGII+BGF, 

that  is,  GHD=zBGF, 

Now,  this  equality  cannot  subsist  unless  BG  and  HD 
run  in  the  same  direction,  viz.,  unless  they  are  parallel. 
Q.  E.  D. 

Cor.  If  the  alternate  angles  between  two  lines  are 
equal,  the  two  lines  are  parallel.  For,  the  alternate  an- 
gles cannot  be  equal  unless  GHD  =  BGF. 


PKOPOSITIOIS'  Y. 

8.  The  sum  of  the  three  angles  of  any  triangle  is 
equal  to  two  right  angles. 

Proof  Let  ABC  be  a  tri- 
angle. Produce  AB  to  D, 
and  from  B  draw  BF  paral- 
lel to  AC.  Then  (ISTo.  6)  the 
angles  FBD  and  CAB  will 


GEOMETRY.  13 


be  equal,  as  also  QBE  and  AGB,  which  are  alternate 
angles.     Hence 

CAB-\-AGB^  CBA  =  EBD^  GBE^  CBA; 
and,  as  the  second  member  of  this  equation  is  equal  to 
two  right  angles  (Ko.  4),  it  follows  that  the  first  mem- 
ber, too,  that  s,  the  sum  of  the  angles  of  the  triangle,  is 
equal  to  two  right  angles.     Q.  E.  D. 

Cot.  1.  If  any  side  of  a  triangle  be  produced,  the  ex- 
terior angle  is  equal  to  the  sum  of  the  two  interior  oppo- 
site.    Thus 

CBD=CAB+ACB. 

Cor,  II.  If  the  sum  of  two  angles  in  one  triangle  is 
equal  to  the  sum  of  two  angles  in  another  triangle,  the 
third  angles  will  also  be  equal  each  to  each. 

Cor.  III.  If  one  angle  of  a  triangle  be  a  right  angle, 
the  sum  of  the  other  two  will  also  be  equal  to  a  right 
angle. 

Cor.  lY.  All  the  angles  of  a  triangle  may  be  acute ; 
but  no  triangle  can  have  more  than  one  right  angle,  or 
more  than  one  obtuse  angle. 


PEOPOSITIOiq"  YI. 

9,  If  two  angles  have  their  sides  parallel,  the  two  an- 
gles will  he  either  equal  or  sujpjplementary . 

Proof.  Let  ^6^  be  parallel 
to  BD^  and  AH  parallel  to 
BF.  Then  we  have  (]S"o.  6) 
HAC=FEC=FBD;  that  is, 
the  angle  FBD  is  equal  to  the 
angle  HA  C.  Prolonging  FB 
to  G,  the  angle  DBG  will  be 
supplementary  to  the  angle 
DBF,  or  to  its  equal  CAH; 


14  ELEMENTS  OF 


for,  two  angles  are  called  supplementary  when  tlieir 
sum  equals  two  right  angles,  as  is  the  case^ith  the 
angles  DBE  and  DB  G.    Q.  E.  D. 


PEOPOSITION  YII. 

10,  If  two  cmgles  have  their  sides  respectively  jper- 
pendicular^  the  two  angles  will  l)e  either  equal  or  sup- 
plementary. 

Proof.  Let  the  angles  BAG  and  GDB  have  their 
sides  respectively  perpendicular.  Producing  GD  till  it 
meets  the  side  AB  in  q 

jEJ  we  shall  have  two 
right-angled  triangles, 
AGE  and  BDE,  in 
the  first  of  which  the 

sum   AEG-\-  GAE  is   ^  B  E 

equal  to  a  right  angle  (No.  8,  Cor.  III.),  and  in  the  sec- 
ond the  sum  BED  -f-  BDE  is  also  equal  to  a  right  an- 
gle ;  and  therefore 

AEG-^  GAE=  BED  +  BDE. 

But  AEG  and  BED  are  one  and  the  same  angle  ;  hence, 
by  reduction,  GAE  =  BDE.  The  angle  BDG  is  sup- 
plementary to  the  angle  BDE.  Hence  it  is  supplemen- 
tary also  to  the  angle  GAE.    Q. E. D. 


PKOPOSITION  YIII.   • 

11.  Two  triangles  which  have  two  sides,  and  the  in- 
cluded angle  of  the  one  equal  to  two  sides  and  the  in- 
cluded angle  of  the  other,  each  to  each,  are  equal  in  all 
their  parts. 


GEOMETRY,  15 


Proof.  In  the  triangles  ABC  and  DEF,  let  AB  = 
DE,  AC=DF,  and  A=:D.  Let  the  triangle  DEF  be 
placed  upon  ABC  ^o  that  the  side  JDE  shall  coincide 
with  its  equal  AB.  Then, 
since  I)=A^  the  side  i>i^will 
lie  in  the  direction  of  A  C ;  and 
since  DF—AC^  the  point  F 
will  coincide  with  the  point  (7, 
and  consequently  the  side  EF  will  coincide  with  the  side 
BC  The  triangles  therefore  coincide  throughout :  which 
means,  in  other  words,  that  they  are  equal  in  all  their 
parts.     Q.  E.  D. 


PEOPOSITIOK  IX. 

12.  Two  triangles  which  have  two  angles^  and  the  in- 
cluded side  of  the  one  equal  to  two  angles  and  the  in- 
cluded side  of  the  other,  each  to  each,  are  equal  in  all 
their  parts. 

Proof  In  the  triangles  ABC  and  DEF,  let  A=D, 
B  =  E,  and  AB  —  DE.  Let  the  triangle  DEF  be 
brought  upon  J.^ (7  so  that  the  side  ^Z>  shall  coincide 
with  its  equal  AB,  Then,  since  Z>=:^,  the  side  DF 
will  lie  in  the  direction  of  A  C j  and  since  E=:  B,  the 
side  EF  will  lie  in  the  direction  of  BC.  The  triangles 
therefore  coincide  throughout,  and  are  equal  in  all  their 
parts.     Q.  E.  D. 


PEOPOSITIOK  X. 

13.  In  every  triangle,  one  side  is  less  than  the  sum 
of  the  other  two  sides. 

Proof  Any  side  J.  (7  of  a  triangle  ABC  exhibits  the 


16  ELEMENTS  OF 


shortest  patli  from  the  point  A  to  the  point  6'/  for  A  C, 
being  a  straight  line,  leads  di- 
rectly from  A  to  C.  It  follows 
that  the  crooked  path  composed 
of  the  sides  AB-^BC  is  longer 
than  J.  6^.     Q.E.D.  A  0 

Cor.  From  the  inequality  AC <AB-\-BC^e  obtain 
AC-BC<AB; 

and  therefore  the  difference  hetween  any  two  sides  of  a 
triangle  is  always  less  than  the  third  side. 


propositio:n"  xl 

14.  If  two  sides  of  a  triangle  are  equal,  the  angles 
opposite  to  them  are  also  equal,  and  the  triangle  is  styled 


Proof  Let  JL^C' be  a  triangle, 
in  which  AB  —  AC.  Bisect  the 
base  BC  2X  D,  and  draw  the  line 
AD.  The  triangle  ABC  will 
thus  be  divided  into  two  trian- 
gles ABB  and  A  CD,  which  will 
be  equal,  as  they  have  equal  sides 
(JS'o.  11)  ;  and  the  angles  B  and  C,  opposite  to  the  com- 
mon side  AD,  will  be  equal.     Q.  E.  D. 

Cor.  The  line  drawn  from  the  vertex  of  an  isosceles 
triangle  to  the  middle  of  the  base  is  perpendicular  to 
the  base.     For  ADB=ADG. 

15.  Conversely,  if  two  angles  of  a  triangle  a/re  equal, 
the  sides  opposite  to  them  are  also  equal,  and  the  trian- 
gle is  isosceles. 

Proof.  In  the  triangle  ABC,  let  B^=  C.  Drawing 
^D  perpendicular  to  BC,  the  two  triangles  ABD  and 


aEOMETBY.  17 


AGD  will  be  equal;  for  they  will  have  all  their  angles 
respectively  equal,  and  the  side  AD  common  (No.  12). 
Hence  the  sides  AB  and  AC  must  be  equal. 

Qor.  L  The  line  drawn  from  the  vertex  of  an  isosceles 
triangle  perpendicularly  to  the  base,  bisects  the  base. 

Cor.  II.  Only  one  perpendicular  can  be  drawn  from 
a  given  point  J.  to  a  given  line  BG.  For  this  line  can- 
not be  bisected  in  more  than  one  point  D. 

Cor.  Ill  Only  one  perpendicular  can  be  raised  from 
any  point  i>  of  a  given  line  BC.  For  this  perpen- 
dicular must  pass  through  the  vertex  A  of  an  isosceles 
triangle  whose  base  is  bisected  in  D,  and  there  can  be 
but  one  direction  from  D  to  A. 

Cor.  lY.  Any  point  in  a  perpendicular  drawn  from 
the  middle  of  a  line  is  at  equal  distances  from  the  two 
extremities  of  the  line.  For  any  such  point  can  be  the 
vertex  of  an  isosceles  triangle  of  which  that  line  will  be 
the  base. 


PKOPOSITIOK    XIL 

16.  In  every  triangle  the  greater  angle  is  o^osite 
the  greater  side. 

Proof.  In  the  triangle  ABC,  let 
AC>AB.  On  the  greater  side  AC, 
take  AD  =  AB,  and  draw  BD,  The 
triangle  ABD  will  be  isosceles,  and 
consequently  the  angles  ABD  and  b< 
ADB  will  be  equal.  But  ADB  = 
DCB  +  DBC  (JSTo.  8,  Cor.  I.);  there- 
fore also  ABD  =  DCB  +  DBC;  that 
is,  ABD  >DCB,  and  a  fortiori  ABC 
>DCB.    Q.E.D. 

Cor,  In  every  triangle  the  greater  side  is  opposite  to 


18  ELEMENTS  OF 


the  greater  angle.  For  in  the  triangle -^^(7,  if  ^  >  (7, 
the  side  A  C  cannot  be  equal  to  the  side  AB,  as  the  tri- 
angle is  not  isosceles  ;  and  ^(7  cannot  be  less  than  AB, 
because  then  B  should  be  less  than  (7,  as  we  have  joist 
shown.    Therefore  AC  >  AB. 


PKOPOSITIOlSr  XIII. 

17.   Two  triangles  having  their   sides   respectively 
equal,  are  equal  in  all  their  'parts. 

Proof,  Let  J.^C'and  J.(7i>  be  two  triangles  having 
the  sides  AB=zAI),  BO=z 
DC,  and  AC=AC.  Join 
them  together  by  their  longest 
sides  A  C,  and  draw  BJ).  The 
triangles  ABB  and  BBC  will 
both  be  isosceles.  Hence  the 
angles  BBCmd  BBC  will  he 
equal,  as  also  the  angles  BCA  and  DC  A,  Hence  also 
the  angles  J.^C'and  ADC  will  be  equal  (ISTo.  8).  And 
therefore  the  two  triangles  are  equal  in  all  their  parts. 
Q.  E.  D. 


PKOPOSITIO:tT  XIY. 

18.  If  two  triangles  have  two  sides  respectively  equal, 
and  the  included  angles  unequal,  the  third  sides  will  he 
unequal :  viz.,  the  greater  side  will  he  ojpposite  the  great- 
er a/ngle. 

Proof  Let  the  two  triangles  ^^6^  and  DBC\i2iYe  two 
sides  respectively  equal,  as  AB=^BDy  and  BC  =  BC; 


GEOMETRY. 


19 


and  let  the  angle  ABC 
be  greater  than  the  angle 
DBG.  The  two  triangles 
being  placed  so  that  their 
shortest  sides  BG  shall 
coincide,  join  AD ;  and 
from  B  draw  BE  per- 
pendicular to  AD^  and 
finally  from  the  point  F^ 

where  BE  intersects  AG^  draw  FD.  The  triangles 
ABD  and  AFD  will  be  isosceles,  because  ABz=.BD 
and  AF=FD. 

ISTow,  A  G=  AF+  FG=  FD + EG.   But  FD  +FG 
>  GD  (No.  13).    Therefore  AG>GD.    Q.  E.  D. 


PKOPOsiTioj^  xy. 

19.  TTie  shortest  line  from  ajpoint  to  a  straight  line 
is  the  ^perpendicular  from  thatjpoint  to  the  line. 

Proof  Let  (7  be  a  point,  and  AB  a  straight  line. 
Draw  GD  perpendicular,  and 
GE  oblique,  to  AB.  In  the 
resulting  triangle  EGD,  GE 
is  greater  than  GD.,  because  it 
is  opposite  a  greater  angle 
(No.  16);  and  this  will  be 
true  also  of  any  other  oblique 
line,  as  GE,  GG,  etc.  Hence  the  perpendicular  GD  is 
the  shortest  line  that  can  be  drawn  from  G  to  the  straight 
line  AB.    Q.  E.  D. 

Gor.  I.  The  perpendicular  is  the  true  distance  from  a 
point  to  a  straight  line. 


20  ELEMENTS  OF 


Cor,  II.  The  oblique  lines  are  greater  or  smaller  ac- 
cording to  their  greater  or  lesser  obliquity.  Thus  CG  > 
CF^  as  the  angle  CFG  opposite  to  CG  is  greater  than 
the  angle  CGF  opposite  to  CF. 

Cor.  Ill  Lines  of  equal  obliquity  have  equal  length, 
and  are  sides  of  isosceles  triangles.  They  meet  the  line 
to  which  they  are  drawn,  at  equal  distances  from  the  foot 
of  the  perpendicular. 

Cor.  I V.  From  a  point  to  a  line,  and  also  from  a  line 
to  a  point,  only  one  perpendicular  can  be  drawn.  For 
there  is  only  one  shortest  line. 


PKOPOSITIOK  XYI. 

20,  In  any  jpolygon,  the  sum  of  the  interior  angles  is 
equal  to  twice  as  many  right  angles,  less  four,  as  the  fig- 
ure has  sides. 

Proof  Let  ABCDE  be  any 
polygon.  From  any  vertex  A 
draw  diagonals,  as  AC,  AD ; 
the  polygon  will  thus  be  divided 
into  triangles,  every  one  of  which 
will  contain  two  right  angles.  If 
the  polygon  has  n  sides,  the  tri- 
angles will  be  71  —  2 ;  hence  the 
sum  of  all  the  angles  will  be  2  (ti  —  2),  or  2/i  —  4. 

Another  proof.  From  a  point 
0  within  the  polygon,  draw  OA, 
OB,  DC,  etc.  The  polygon  will 
be  divided  into  as  many  trian- 
gles as  it  has  sides.  The  sum  of 
all  the  angles  will  therefore  be 
equal  to  twice  as  many  right  an- 
gles as  there  are  sides.     But  the 


CfEOMETRY,  51 


angles  about  0,  whose  sum  is  equal  to  four  riglit  angles, 
do  not  belong  to  the  polygon.  Hence,  taking  them 
away,  the  angles  of  the  polygon  will  amount  to  twice 
as  many  right  angles,  less  four,  as  the  figure  has  sides. 
Q.  E.  D. 


PROPOSITIOlSr  XYII. 

21.7/^  the  opposite  sides  of  a  quadrilateral  are  equal, 
they  are  parallel,  and  the  figure  is  a  parallelogram. 

Proof.  Let  AB  CD  he  a  quadrilateral,  in  which  AB=z 
CD,  and  AD  =  BO.  Draw  the  diagonal  AC;  the  tri- 
angles A  CD  and  A  CD,  as  having 
equal  sides,  will  be  equal  in  all 
their  parts  (No.  17).  Hence  the 
angles  DAC  and  ^6^J.,  opposite 
to  the  equal  sides  DC  and  AD, 
will  be  equal,  and  therefore  the 
sides  AD  and  DC  will  be  parallel  (No.  7,  Cor.)  Simi- 
larly, the  angles  A  CD  and  DA  C,  opposite  to  the  equal 
sides  AD  and  DC,  will  be  equal,  and  therefore  the  sides 
AD  and  DC  will  be  parallel.     Q.  E.  D. 

Cor.  I.  In  all  parallelograms,  the  opposite  sides,  as 
also  the  opposite  angles,  are  equal.  For,  the  diagonal 
will  make  equal  alternate  angles  between  the  parallel 
sides ;  and  consequently  will  divide  the  figure  into  two 
equal  triangles,  in  which  the  equal  sides  will  be  parallel 
and  opposite. 

Cor.  II.  If  two  parallel  lines  intercept  two  other  par- 
allel lines,  they  form  a  parallelogram. 

Cor.  III.  The  diagonals  of  a  parallelogram  bisect 
each  other.      For,  if  AD  CD  is    a  parallelogram,  the 


22 


ELEMENTS  OF 


diagonals  make  the  alternate  angles  ABD 
DC  A— CAB.      The    trian- 
gles AOB   and   DOC  have 
therefore    two    equal    angles  / 

about   the    equal    sides  AB        /       .^.-•••-  0\ 

and  DC^    and    consequently      /--•••-" X^ 

they  are  equal  in  all  their  ^ 
parts.  Hence  AO—OC  2.-^^  BO—  OD. 
Cor.  IV.  The  diagonals  of  a  rhombus 
bisect  each  other  at  right  angles.  For 
the  rhombus  is  a  parallelogram  whose 
sides  are  all  equal ;  whence  it  follows 
that  the  two  diagonals  are  the  bases  of 
isosceles  triangles.  Hence  A  C,  which  bi- 
sects BD^  is  perpendicular  to  BD  (No. 
14,  Cor.) 


BDC,  and 

B 


GEOMETRY, 


23 


BOOK   II 


RECTILINEAR  AREAS,  AND  RELATIONS 
OF  SIMILAR  FIGURES. 

23.  In  measuring  surfaces,  the  unit  of  measure  is  a 
square  having  for  its  side  the  unit  of  length.  This  unit 
is  arbitrary.  Let  the  square  a^  be  our  unit  of  surface. 
Then  any  rectangular  surface  ABGD  will  be  measured 
by  ascertaining  how  many  times 
the  surface  a?  is  contained  in  it. 
If,  for  instance,  the  base  AB  is 
equal  io  five  units  of  length,  and 
the  altitude  AD  is  equal  to  six 
of  the  same  units,  then  the  sur-     .  . 

face  ABGD  will  contain  six  I  I  I  I  I  I  I  **' 
times  a  row  of  fiA)e  square  units,  ' 
that  is,  30  units  of  surface,  which  are  obtained  by  mak- 
ing the  product  of  the  base  AB  into  the  altitude  BC ; 
and  then  we  say  that  the  area  of  ABGD,  as  measured 
by  such  a  unit,  is  =  30. 

We  have  supposed  AB  and  BG  to  contain  the  unit  of 
length  a  an  entire  number  of  times.  Let  now  m  and  n 
be  any  two  numbers,  and  let  us  assume  AB=zma, 
BG:=zna,    Then 

ABxBG=mna\ 


and 


AB      BG 


m 


X 


n 


ITow,  it  is  plain  that  these  equations  must  be  true, 
whether  m  and  n  are  entire  or  fractional,  and  whether 
they  are  commensurable  or  incommensurable  with  a. 
For  we  have  in  all  cases 


24 


ELEMENTS  OF 


AB 

m 


ma 

TYh 


BC 


n 


na 
n 


Hence  the  area  of  a  rectangle  is  always  expressed  by  the 
product  of  its  base  into  its  altitude,  whatever  may  be 
their  relation  to  the  unit  of  measure. 

Surfaces  that  are  not  rectangular  can  also  be  measured 
by  any  square  unit,  as  we  are  going  to  show  in  this  sec- 
ond book. 


PEOPOSITIOJST  I. 

23.  Parallelograms  having  equal  bases  between  the 
same  parallels  have  equal  areas. 

Proof.  Let  ABCD  be  a  parallelogram.  Prolong  CD 
towards  E^  and  construct  on 
AB  another  parallelogram, 
ABFE,  by  drawing  any  two 
parallel  lines  AE  and  BF. 
The  two  parallelograms  will 
have  equal  areas. 

For  the  area  of  ABCD  is 
equal  to  the  area  of  the  whole  figure  ABDE  minus  that 
of  the  triangle  A  CE\  and  the  area  of  ABFE  is  equal 
to  the  area  of  the  whole  figure  ABDE  minus  that  of 
the  triangle  BED.  J^ow,  the  triangles  ACE  and  BED 
are  equal  (No.  11).  Therefore  the  expression  of  the 
area  of  the  two  parallelograms  is  exactly  the  same ;  and 
consequently  those  two  areas  are  equal.     Q.  E.  D. 

Cor.  1.  The  area  of  a  paral- 
lelogram is  equal  to  the  base 
multiplied  by  the  perpendicular 
height.  For  construct  upon  the 
same  base  and  between  the  same 
parallels  a  rectangular  parallelo- 


GEOMETRY,  25 


gram  ABFE ;  the  areas  ABCD  and  ABEF  will  be 
equal.  But  ABFE=  AB  X  BF  (No.  22) ;  therefore 
ABDG^ABy^BFz^ABy^DH. 

Got.  II.  As  the  diagonal  divides  the  parallelogram 
into  two  equal  triangles  (Ko.  21, 
Cor.  I.),  the  area  of  a  triangle  is  ^ 

equal  to  one-half  the  area  of  the        y^  \ 
parallelogram,  having    the    saine    x  \ 

hase  and  the  same  height.     Thus  ^    ^ 

area  ABG^  \ABx  GD. 

Gor.  111.  Triangles  having  equal  bases  and  equal 
heights  have  also  equal  areas  ^  for  the  expression  of 
both  areas  is  the  same. 

Gor.  1 V.  The  area  of  a  trapezoid  (a  quadrilateral,  of 
which    two    sides    are    parallel) 

equals  one-half  the  sum  of  its  P ^5 

parallel  sides  multijplied  hy  their        /         ^^      i\ 
distance.       For  if  ABGD  is  a       /  ^■■■■^  \  \ 

trapezoid,  it  may  be  divided  by  a     ^ H — b 

diagonal  A  G  into  two  triangles, 

J.^(7and  ADG,  having  parallel  bases  AB  and  i>(7and 
a  common  altitude  GH.  Hence  the  area  of  the  trape- 
zoid will  be  expressed  by 

\ABy.  GH-^  \  GDx  GH,  or  GH  (^^+^"^)» 


PEOPOSITIOIsr  II. 

34.  The  square  described  on  the  sum  of  two  lines  is 
equal  to  the  sum  of  the  squares  m^ade  on  each  line,  jplus 
twice  the  rectangle  contained  by  the  two  lines. 

Proof  Let  AB  and  ^6^  be  the  two  lines,  and  A  GDF 


26 


ELEMENTS  OF 


the  square  described  on  their  sum  A  C.     Draw  BE  par- 
allel to  CD,  and  complete  the  square 
ABHG  by  drawing  GI  parallel  to    ^  ED 

A  G.  The  area  A  CDF  will  thus  be 
the  sum  of  four  areas,  viz. :  ABGII  G 
=  AB\  HEFG^GHy^EH^AB 
xBC,EIBC^HBxBCz=^ABx 
BG,  and  EIDE=:zBG\  Conse- 
quently the  square  of  the  sum  AB 
+  BGis> 

AB''-\-'iABxBG\BC\    Q.E.D. 


H 

peopositio:n  hi. 

25.  The  square  described  on  the  difference  of  two 
lines  is  equal  to  the  sum  of  the  squares  made  on  each 
line,  minus  twice  the  rectangle  contained  hy  the  two 
lines. 

Proof  Let  AB  and  BGh^  the  two  lines,  and  A  CDE 
the  square  described  on  their 
difference  AG.  Construct  on 
AB  the  square  ABFG,  and  on 
BG  the  square  GBIK,  and  pro- 
duce ED  to  H.  The  area  of 
the  square  A  GDE  will  be  equal 
to  the  area  of  the  whole  figure, 
minus  the  two  areas  GEJIFaind 
DHIK.  JS'ow,  the  area  of  the 
whole  figure  is  ABFG-\-  GBIK, 
or  AR^BG\  and  the  two 
areas    GEHF  and  DHIK  are 


P 

A  C' 


GEOMETRY.  2l 


both  equal  to  ABxBG.     Therefore  the  square  of  the 
difference  AB  —  BG  is 

AB'-2ABxB0+Ba\    Q.E.D. 


PKOPOSITIO]^  lY. 

36.  The  difference  of  the  squares  described  on  any 
two  lines  is  equal  to  the  rectangle  of  their  sum  hy  their 
difference. 

Proof  Let  AB  and-  AG  h^  the 
two  lines,  and  ABBE,  A  GHF,  their 
squares.  The  difference  between 
these  two  squares  is  the  sum  of  the 
two  rectangles  FGDE  and  GBGH. 
1^0 w,  FGDE=^FG  x  FE=ABx 
BG,  and  GBGH^GHxBG^AG 
XBG.     Hence  their  sum  is 

ABxBG+AGxBG=::{AB-\-AG)BG, 

which,  because  BG=zAB—AG,  is  the  same  as  {AB -\- 
A G)  {AB  -AG)\  and  therefore 

AB'-  AG'^  {AB  +  AG)  {AB  -  A G).    Q.  E.  D. 


PKOPOSITIOE"  Y. 

2*7.  The  squa/re  made  on  the  hypotenuse  of  a  right- 
angled  triangle  equals  the  sum  of  the  squares  made  on 
the  other  two  sides. 

Proof.  By  the  term  hypotenuse  is  meant  the  side  op- 
posite the  right  angle.     Let,  then,  AB  be  the  hjpote- 


2S 


ELEMENTS  OF 


nuse  of  the  triangle  ABC.  Make  the  squares  ABHK, 
CBDE.ACFG,  and  draw  CL  perpendicular  to  HK, 
Joining  GB  and 

CH^  the  triangles  E 

GAB  and  CAH 
will  be  equal,  as 
thej  have  two 
equal  sides  en- 
closing an  equal 
angle ;  for  the 
angles  GAB  and 
CAH  are  both 
composed  of  a 
right  angle  and 
of  a  common  an- 
gle CAB.  But 
the  triangle 
GAB  has  the 
same    base    and 

the  same  altitude  as  the  square  A  CFG ;  hence  its  area 
is  equal  to  one-half  the  area  of  the  same  square.  On  the 
other  hand,  the  triangle  6ZA^has  the  same  base  and  the 
same  altitude  as  the  rectangle  AILH ;  and  thus  its  area 
is  equal  to  one-half  the  area  of  the  rectangle.  Therefore 
twice  the  triangle  GAB^  that  is,  the  square  A  CFG, 
equals  twice  the  triangle  ACH,  that  is,  the  rectangle 
AILR, 

In  the  same  manner,  joining  AD  and  CK,  it  will  be 
proved  that  the  triangles  ABD  and  CBK  are  equal ; 
also  that  ABD  is  equal  to  one-half  the  square  CBDE, 
and  that  CBK  is  equal  to  one-half  of  the  rectangle 
BKLIy  and  consequently  that  the  square  CBDE  and 
the  rectangle  BKLI  are  equal. 

But  the  two  rectangles  AILH  and  BKLI,  taken  to- 


GEOMETRY.     .  29 


gether,  make  the  square  ABKH,  Therefore  the  squares 
AGFG  and  GBDE^  taken  together,  equal  the  square 
ABKH,    Accordingly, 

AB'z=.AG'^BG\    Q.  E.D. 

Cor.  Two  right-angled  triangles,  in  which  the  hyjpote- 
nuse  and  one  side  are  respectively  equal,  are  equal  in  all 
their  parts.  For  let  the  two  triangles  be  ABC  and 
A'B'C\  AB  and  A'B'  being  the  hypotenuses.  Then 
we  shall  have  in  the  one 

A:B'  =  BC'+  AC' ;  whence  AG'^ AB'- BG\ 
and  in  the  other 


A'B''^  B'C'-^A'G";  whence  A' C''=A'B''-B'C'\ 

If,  then,  the  hypotenuses  AB  and  A^B^  are  equal,  and 
the  sides  BC  and  B^G^  are  also  equal,  the  third  sides 
A^G^  and  AC  must  also  be  equal;  which  shows  that 
the  two  triangles  are  equal  (No.  lY). 


PROPOSITIOI^  YI. 

2S,  In  any  triangle,  the  square  of  a  side  opposite 
an  acute  angle  equals  the  sum  of  the  squares  of  the  other 
two  sides,  minus  twice  the  ^product  of  one  of  these  sides 
into  the  projection  of  tJie  other  upon  its  direction. 

Proof.  Let  ABC  be  a  triangle,  ^ 

in  which  AC  i^  opposite  the  acute  y\ 

angle  B.     Draw  CD  perpendicu-  y^ 

lar  to  AB.     In  the  right-angled  y^ 

triangle  A  CD,  we  have  (Prop.  Y.)        / 

AC'zz^  AB'+  Cl>\  A 


30  ELEMENTS  OF 


But  (Eo.  25) 

Al>'z=^  {AB-DBf^  AB'-  2AB  x  DB-\-DB\ 
and  (Ko.  27,  Cor.)  CD''^  CB'-  I)B\ 
Hence,  substituting  and  reducing, 

AG^=AB''-\-  CB'-  "iABxBB; 
DB  being  the  so-called  jprojection  oi  BG  upon  AB, 


Q.  E.  D. 


PEOPOsiTioN  yn. 


^9.  In  any  obtuse-angled  triangle^  the  square  of  the 
side  opposite  the  obtuse  angle  equals  the  sum  of  the 
squares  of  the  other  two  sides ^  plus  twice  the  prodxtct 
of  one  of  these  sides  into  the  projection  of  the  other  upon 
its  direction. 

Proof    Let  ^^6^  be  a  C 

triangle,  in  which  AG  i^     '  ^ 

opposite  the  obtuse  angle  ^^  / 

B.    Draw  GD  perpendicu-  ^^       / 

lar  to  AB  produced.     In        ^^  / 

the    right-angled     triangle    '^ 1. 

AGD^  we  have 

AG"  z::^  AD'' -\- GD\ 

But  AD''  =3  {AB-\-BDY  =  AB'-\-  2AB  x  BD+BD% 

and  GD'=GB'-BD%' 

hence,  substituting  and  reducing, 

AG'z=:AB'+  GB'+2ABxBD; 

BD  being  the  projection  of  the  side  BG  on  the  direc- 
tion of  the  other  side  AB.     Q.  E.  D. 


GEOMETRY. 


31 


PEOPOSITIOlNr  YIII. 

30.  In  any  triangle^  the  sum  of  the  squares  described 
on  two  sides  is  equal  to  twice  the  square  of  one-half  the 
third  side^  plus  twice  the  square  of  the  line  drawn  to  the 
middle  jpoint  of  that  side  from  the  vertex  of  the  ojpjposite 
angle. 

Proof.  In  the  triangle 
ABC  bisect  AB  at  E  and 
join  CE.  Draw  also  GH 
perpendicular  to  AB.  Then 
in  the  triangle  ACE  we 
shall  have  (JSTo.  29) 

AG'^AE'^CE' 

-\-2AExEH, 

and  in  the  triangle  BCE 
we  shall  have  (E"o.  28) 

BG'=:  BE'+  CE'-  2BEx  EH. 

Adding  together  these  two  equations,  and  observing  that 
AE  and  BE  are  equal,  we  shall  have 

AC^  +  BG''  =  2AE'  +  2(7X^    Q.  E.  D. 

Cor.  The  sum  of  the  squares  of  the  four  sides  of  a 
jparallelogram  is  equal  to  the  sum  of  the  squares  of  its 
diagonals.  For  drawing  AD  and  BD  parallel  to  BG 
and  A  G  respectively,  and  prolonging  CE  to  the  point 
Z>,  the  figure  will  be  a  parallelogram,  of  which  AB  and 
CD  are  the  diagonals.  Now,  the  equation  just  found, 
multiplied  by  2,  gives 


32  ELEMENTS  OF 


wliicli  may  be  written  thus, 

AO'+£D'+BC'+A^'=  AB'+  CD\ 


PEOPOSITIOI^  IX. 

31,  Parallelograms  Jiamng  equal  altitudes  are  to  one 
another  as  their  hases  /  and,  if  they  have  equal  bases, 
are  to  one  another  as  their  altitudes. 

Proof.  Let  P  designate  the  area  of  a  parallelogram 
having  a  base  h  and  an  altitude  h.  Let  also  P'  designate 
the  area  of  another  parallelogram  having  a  base  V  and 
an  altitude  A'.     We  shall  have  (:N'o.  23,  Cor.  I.) 

P:=hXh,        P'^l'y.h'; 

hence  P\  P::hxh\  h'xh'. 

If  A  =  A^then  P'.P'wl'.y, 

and  if  l-=^V,  then   P  \P'\\h\h'.     Q.  E.  D. 

Cor.  Triangles  having  equal  altitudes  are  to  each 
other  as  their  hases  /  and  triangles  homing  equal  hases 
are  to  each  other  as  their  altitudes.  For  triangles  are 
semi-parallelograms,  and 

\P:^P:'.P'.P'. 


POEPOSITION  X. 

32.    In  any  triangle,  a  line  parallel  to  one  of  the 
sides  divides  the  other  sides  into  ^proportional  jparts. 

Proof.  In  the  triangle  ABC,  draw  PE  parallel  to 


GEOMETRY.  33 


BC.  Draw  BE,  The  triangles 
BDE  and  DAE,  with  a  common 
vertex  E,  will  have  the  same  alti- 
tude; hence  they  will  be  propor- 
tional to  their  bases  (No.  31,  Cor.) 
Therefore 

BDE:  DAE::  BD  :  AD. 

]N"ow    draw    DC.       The    triangles 
ODE  and  DAE,  with  a  common     ^ 
vertex  Z>,  will  have  the  same  alti- 
tude;  hence  they,  too,   will  be  proportional    to  their 
bases.     Therefore 

DOE:  DAE::  GE\EA, 

But,  as  DE  and  BG  are  parallel,  the  triangles  BDE 
and  DOE,  which  have  a  common  base  DE,  have  also 
the  same  altitude,  viz.,  they  are  equal.  Hence,  in  the 
two  proportions  just  found,  the  first  ratios  are  equal. 
Consequently  th  e  last  ratios  also  are  equal,  viz. : 

BD  :  AD  ::  CE :  EA.    Q.E.D. 

33.  Conversely,  If  a  line  divides  two  sides  of  a 
triangle  into  proportional  parts,  that  line  is  paral- 
lel to  the  third  side  of  the  triangle.  For  the  two  ratios 
BD  :  AD  and  CE :  EA  cannot  be  equal,  nor  can  they 
form  a  proportion,  unless  the  ratios  BDE  :  DAE  and 
DC^:  DAE,  with  which  they  are  strictly  connected, 
be  also  equal,  that  is,  unless  the  triangles  BDE  and 
DCB  be  equal.  Now,  these  triangles  have  a  common 
base  DE  I  they  must  therefore  have  an  equal  altitude ; 
and  consequently  their  vertices  B  and  C  must  be  on  a 
line  parallel  to  their  base  DE, 


34  ELEMENTS  OF  \ 


PEOPOSITIOlSr  XL  /^' 

34.  Similar  (that  is,  equiangular)  triangles  have  their 
corresjponding  sides  proportional. 

Proof.  In  tlie  triangle  ABC,  draw  DE  parallel  to 
AB.     Then   the   triangle    CDE 
will  be   equiangular  with  ABC, 
and  we  shall  have 

AD  \  DC  w  EB  \  EC, 
whence,  bj  composition, 
AD-]-DC\DC:\EB+EC\  EC, 
that  is, 

AC'.DCwBC'.EC.      (1) 

If  we  now  draw  EF  parallel  to  A  C,  we  shall  have 
also 

BF  :FA::BE  :EC, 

whence,  by  composition, 

BF+FA  :FA::  BE+EC :  EC, 

that  is,  BAiFA  ::  BC  :  EC, 

or,  because  AF=  DE, 

BA:DE::BC  :EC  (2) 

Hence,  comparing  the  two  proportions  (1)  and  (2), 

AC:DC::BC  :EC::BA  :  DE. 

Therefore  the  corresponding  sides  of  any  two  equiangu- 
lar triangles  ABC  and  DEC  are  proportional.  Q.  E.  D. 
J^ote.  The  corresponding  sides  are  those  which  are  op- 
posite to  equal  angles,  and  are  called  homologous,  be- 
cause they  form  equal  ratios. 


GEOMETRY. 


35 


35.  Conversely,  Triangles  whose  sides  are  jproj^or- 
tional  are  equiangular  and  therefore  similar.  For  let 
there  be  between  the  sides  of  two  triangles  ABC  and 
abo  the  proportion 

AB:ab::AG'.aG::BC\hG. 

Taking  on  ^^  a  length  AD 
=:ab,  and  on  AG  ?i  length 
AE—  ac,  and  drawing  DE, 
we  shall  have 

AB  :  AD  ::  AC  :  AE. 

Now,  the  existence  of  this 
proportion  requires  that  DE 
be  parallel  to  BG  (No.  33).  Therefore  DE  and  BG  are 
parallel,  and  the  triangles  ABG  and  ADE  are  equian- 
gular.    Hence  we  have  also 

AG'.AEwBG'.DE, 
that  is,  AG\  ao  ::  BG  :  DE. 

But  we  have  already 

AG  :  ac  ::  BG  :  ho ; 
therefore  DE=  ho. 

Thus  the  three  sides  of  ADE  are  respectively  equal 
to  the  three  sides  of  aho;  and  therefore  these  two  trian- 
gles are  equal  in  all  their  parts.  But  ADE  and  ABG 
are  equiangular.  Therefore  abo  and  ABG,  too,  are 
equiangular.     Q.  E.  D. 

Gor.  I.  Triangles  whose  sides  are  respectively  paral- 
lel are  equiangular  and  similar. 

Gor.  II.  Triangles  having  an  equal  angle  between 
two  proportional  sides  are  equiangular  and  similar. 


36 


ELEMENTS  OF 


PEOPOSITION^   XII. 

36.  Triangles  whose  sides  are  respectively  perpendi- 
cular a/re  equiangular  and  similar. 

Proof.  Let  ABC  and  DEF  have  tlieir  sides  mutual- 
ly perpendicular,  that  is,  DE  perpendicular  to  AB^  EF 
to  BC,2indiJDF to  AC.  Pro- 
long DE,  EF,  ED  till  they 
meet  AB,  BC,  and  AC  re- 
spectively. Then  we  shall 
have  three  quadrilaterals 
AHDG,  HEIB,  FICG,  in 
each  of  which  two  angles 
will  be  right  and  two  other 
angles  will  be  supplementary ; 
for  the  sum  of  the  four  an- 
gles must  be  equal  to  four 
right  angles  (No.  20).     Thus 

we  have  GAH -\-  GDH—  two  right  angles;  but  we 
have  also  (Ko.  4)  EDF-\- GDH  =  two  right  angles. 
Hence  GAH=  EDF. 

And,  in  the  same  manner,  it  will  be  found  that  HBI 
is  equal  to  DEF,  and  ICG  equal  to  DEE.    Q.  E.  D. 


PEOPOSITIOlSr  XIIL 

37.  In  any  triangle,  the  line  that  hi  sects  one  angle 
divides  the  opposite  side  into  parts  proportional  to  the 
adjacent  sides. 

Proof.  Let  ABC  be  any  triangle.  Draw  CD  bisect- 
ing the  angle  (7,  and  produce  J.  67  by  a  length  CE=  CB, 


GEOMETRY. 


37 


Draw  EB ;  the  triangle  EGB  will  be  isosceles,  viz.,  the 
angles  QBE  and  CEB  will  be  equal.  Now,  the  exte- 
rior angle  ACB  is  equal  to 
the  sum  of  the  two  angles 
CBE  and  CEB  Q^o,  8, 
Cor.  I.);  therefore  ACB,  or 
2DGB,  is  equal  to  2  CBE, 
and  DCB=z  CBE.  But 
these  last  two  angles  are  al- 
ternate angles  between    CD 

and  BE.  Hence  CD  and  BE  are  parallel  (ISTo.  7,  Cor.), 
and  the  triangles  ACD  and  AEB  are  similar;  and 
therefore  ^i>  :  DB  ::  AC :  CE ;  or,  since  CE  and  (7^ 
are  equal  by  construction, 

AD'.DBwAC  \  CB.    Q.E.D. 


PKOPOSITION  XIY. 

38.    Triangles  which  have  one  angle  equal  are  to 
each  other  as  the  rectangles  of  the  including  sides. 

Proof.  Let  the  triangles  ABC  and  CDE  have  one 
equal  angle  (7,  and  conceive  them  to  be  so  placed  that 
the  equal  angles  ACB  and 
DCE  shall  be  vertically  op- 
posite. Then  join  AE.  .The 
triangles  A  CB  and  A  CE  will 
have  a  common  vertex  A,  and 
consequently  (ISTo.  31,  Cor.) 
they  will  be  to  each  other  as 
their  bases.     Thus 

ACB:  ACE::  BC  :  CE. 

The  triangles  ACE  and  ECD  will  also  have  a  common 


38  ELEMENTS  OF 


vertex  JE,  and  consequently  tliej,  too,  will  be  to  eacli 
other  as  their  bases.     Hence 

ACE  \DCE\\AG  :  CD. 

Multiply  these  two  proportions,  term  by  term,  and  can- 
cel the  common  factor  A  CE ;  the  result  will  be 

ABC:  CDE::ACxBC:  CDxCE,    Q.E.D. 


PEOPOSITIOISr    XY. 

39.  In  a  right-angled  triangle^  the  perpendicular 
drawn  from  the  right  angle  to  the  hyjpotenuse^  divides 
the  triangle  into  two  triangles  similar  to  the  given  tri- 
am^gle  and  to  each  other. 

Proof.  Let  ^^6^  be  a  triangle  right-angled  at  (7,  and 
let  CD  be  the  perpendicular  on  the  hypotenuse  AB, 
Designate  the  angle  DCB 
by  a,  and  the  angle  DCA 
by^.  In  the  triangle  J.  j&  (7 
the  sum  A-\-B  of  the 
oblique  angles  is  equal  to 
a  right  angle  (No.  8,  Cor. 
III.);  also  in  the  triangle 
DCB  the  sum  ^  +  a  of  the  oblique  angles  is  equal  to  a 
right  angle ;  and  similarly  in  the  triangle  DCA  the  sum 
^  _|_  ^  of  the  oblique  angles  is  equal  to  a  right  angle. 
We  have  therefore 

-4-f^  =  ^-f  «  =  ^  +  /5; 

whence  J.  =  a,  Bz=z^.  And  thus  it  appears  that  the 
triangles  ABC,  ADC,  BDC  are  mutually  equiangular. 
They  are  therefore  similar  to  one  another.     Q.  E.  D. 


GEOMETRY.  39 


Cor.  I.  Each  side  ahout  the  right  angle  is  a  meanjpro- 
portional  hetween  the  hypotenuse  and  the  adjacent  seg- 
ment. For  in  the  similar  triangles  ABC  and  ACD  we 
have 

AB\AC\\AC\AD, 

and  in  the  similar  triangles  ABC  and  BDC,  we  have 

AB'.BGwBC'.BD. 

Cor.  II.  The  perpendicular  CD  is  a  mean  propor- 
tional hetween  the  segments  of  the  hypotenuse.  For  in 
the  similar  triangles  ADC  and  BDC  we  have 

AD:  CD::  CD  :  DB. 

Cor.  III.  From  Corollary  I.  we  have 

AU'^ABxAD,  and  BC'^ABxBD. 

By  adding  these  two  equations,  we  obtain 

AC'  +  BC'=  AB  {AD  +DB)  =  AB% 

a  result  identical  with  that  obtained  above  (^o.  27). 


PEOPOSITION  XYI. 

40.  Similar  triangles  a/re  to  one  another  as  the  squares 
of  their  homologous  sides. 

Proof.  Let  ABC  and  ^ 

abc  be  any  two   similar  /fC 

triangles.   Draw  CD  and  /     !  \  X 

cd  perpendicular  to  the  /         1    \  /\ 

homologous     sides    AB        /  |      \         /     \\ 

and  ah  respectively.   The    ^^ g ^g  J- ^ — \ 

area   of    ABC   will    be 

■J-J.J&X  CD^  and  the  area  of  ahc  will  be  ^ahxcd  (No.  31, 

Cor.)     Hence 


40 


ELEMENTS  OF 


ABC  :  abo  ::  ABxCD  :  ahxcd. 

But  since  the  triangles  A  CD  and  acd  are  also  similar 
(for  A  =  a,  and  ABO  =  ado,  and  consequently  ACB= 
acd ),  we  have  also 

OB:  cd  ::A0  :  ac  ::  AB  :  ab  ; 

hence,  substituting  AB  :  ah  in  the  place  of  OB  :  cd  in 
the  above  proportion,  we  have 

ABO  lahc::  AB'  :  d>\    Q.  E.  D. 


PKOPOSITION  XYIL 

41,  The  perimeters  of  similar  polygons  are  to  each 
other  as  any  two  homologous  sides  /  and  their  areas  are 
to  each  other  as  the  squares  of  any  two  homologous  sides. 

Proof  Similar  poly- 
gons are  polygons  which 
can  he  divided  into  the 
same  numher  of  simi- 
lar tria/ngles  similarly 
placed.  Hence  all  the 
sides  and  diagonals  of  the 
one   are  proportional   to 

the  corresponding  sides  and  diagonals  of  the  other. 
ABOBE  and  ahcde  be  two  similar  polygons, 
shall  have 


Let 
We 


AB 

ah 


BO 

he 


OB      BE      AE 


cd 


de 


ae 


whence,  by  an  algebraic  rule  regarding  equal  ratios,  we 
obtain 


GEOMETRY.  41 


AB+BC^CI)+DE+EA  ^AB  _BG  _ 

ah  -\-  1)0  -\-  cd  -{-  de  -\-  ea         ah  ho 

But  the  first  member  of  this  equation  is  evidently  the 
ratio  of  the  two  perimeters.     Hence 

Perim.  ABODE:  Perim.  abode::  AB  :  ab::BC\  ho::... 

"We  have,  similarly,  for  the  areas  of  the  triangles  of  each 
polygon, 

ABC _  AGP  _  APE _  AB^  _  BO^  _ 

ahc  aed  ade  ab^  Sc"^ 

Whence,  by  the  algebraic  rule  just  mentioned,  we  have 

ABC -^r  AGP -\- APE  _  AB^  _  BG^  _ 

abo  -\-   acd   +  ade     ~    ^^  bd"   ~  '  '  ' 

But  the  first  member  of  this  equation  is  evidently  the 
ratio  of  the  two  polygonal  areas.     Hence 


AresiABCPE:  2ire^abcde::AB'  :ab'::BG'  :  bo' 


And  these  results  are  equally  true,  whatever  the  number 
of  sides  allotted  to  the  similar  polygons. 


42 


ELEMENTS  OF 


BOOK   III 


THE    CIRCLE    AND     THE    REGULAR 
POLYGONS. 

43,  The  circle  is  a  plane  figure,  bounded  by  a  curved 
line,  every  point  of  which  is  equally  distant  from  a  point 
within,  called  the  centre.  The  bounding  line  is  called 
the  circumference  J  and  any  straight  line  drawn  from 
the  centre  to  a  point  of  the  circumference  is  called  a  ra- 
dius. A  line  drawn  through  the  centre  and  terminating 
at  both  ends  in  the  circumference  is  called  a  diameter. 
An  a/rc  is  any  part  of  the  circumference ;  a  chord  is  a 
straight  line  joining  the  extremities  of  an  arc.  The  part 
of  the  circle  included  between  an  arc  and  its  chord  is 
called  a  segment  of  the  circle ;  whereas  the  part  of  the 
circle  included  between  two  radii  is  called  a  circular  sec- 
tor. Thus,  in  the  annexed 
figure,  0  is  the  centre  of  the 
circle,  DM  a  radius^  BD  a 
diameter^  CN  a  chords  the 
portion  CHN  of  the  circle  a 
segment^  the  portion  DMA  a 
sector. 

43.  The  circumference  of 
a  circle  is  usually  divided 
into  360  equal  parts,  called 
degrees  f  each  such  degree  is 

again  divided  into  60  parts  called  minutes ;  and  each 
minute  subdivided  into  60  parts  called  seconds.     It  fol- 


GEOMETRY. 


43 


lows  that,  if  we  draw  two  diameters  A  C  and  BD  per- 
pendicular to  each  other,  the  arcs  AB,  BC^  CD^  DA 
(which  are  evidently  equal)  will  each  be  equal  to  90  de- 
grees; and  consequently  the  arc  of  90  degrees  corre- 
sponds to,  and  is  taken  as  the  measure  of,  a  right  angle 
at  the  centre  of  the  circle.  So  also  arcs  greater  or  less 
than  90  degrees  correspond  to  angles  greater  or  less  than 
the  right  angle,  and  are  taken  as  their  measure.  Thus, 
if  the  arc  AM  is  equal  to  30  degrees,  the  angle  MO  A, 
to  which  it  corresponds,  is  an  angle  of  30  degrees. 

Degrees,  minutes,  and  seconds  are  distinguished  by 
special  marks :  tlius,  to  designate  30  degrees,  4  minutes, 
and  40  seconds,  we  write  30°  04'  40'^ ;  and  to  designate 
25  minutes  and  50  seconds,  we  write  0°  25'  50''. 


PEOPOSITIOK  I. 

44.  A  radius  perpendicular  to  a  chord,  hisects  the 
chord,  and  also  the  arc  subtended  by  it. 

Proof.  Let  AB  be  a  chord,  and  01)  a  radius  perpen- 
dicular to  it.  Draw  the  radii 
OA  and  OB.  Then  the  right- 
angled  triangles  AOC  and 
BOG  will  have  the  sides  AO 
and  BO  equal,  and  CO  com- 
mon. They  are  therefore  equal 
in  all  their  parts  (^o.  27,  Cor.) 
Hence  AG~BC;  that  is,  the 
chord  is  bisected.     Q.  E.  D. 

Again,  the  angles  AOD  and 
BOD  are  equal,  and  the  sides  AO,  BO,  DO  are  equal ; 
therefore  the  chords  AD  and  BD  are  equal,  as  also  are 


44 


ELEMENTS  OF 


the  arcs  AD  and  BD,  wliicli  correspond  to  the  equal 
angles.     Q.  E.  D. 

Cor.  I.  In  the  same  circle,  equal  arcs  are  subtended  hy 
equal  chords.  For  if  the  arcs  AD  and  DB  are  equal, 
the  angles  AOD  and  BOD  are  also  equal  (No.  43),  and 
the  triangles  ADO  and  BDO  will  be  equal  (No.  11). 
Hence  the  chords  AD  and  BD  will  be  equal. 

Cor.  II.  In  the  same  circle,  equal  chords  suhtend 
equal  arcs,  and  greater  chords  subtend  greater  arcs. 
For  if  the  chords  AD  and  DB  are  equal,  the  angles 
AOD  and  BOD  are  equal,  as  also  the  corresponding 
arcs  ;  and  if  the  chord  AD  were  greater  than  the  chord 
BD,  the  angle  A  OD  would  be  greater  than  the  angle 
BOD  (No.  18) ;  and  consequently  the  arc  AD  would  be 
greater  than  the  arc  BD, 


PEOPOSITION  II. 

45.  In  the  same  circle,  equal  chords  are  equally  dis- 
tant from  the  centre,  and  of  unequal  chords  the  greater 
is  nearer  the  centre. 

Proof  Let  AB  and  ^4^  be 
two  chords.  From  the  centre 
0  draw  the  radius  OA,  as  also 
the  lines  OP,  OQ  respectively 
perpendicular  to  AB  and  A  C. 

Then  we  shall  have 

OP  =  (92'  -  AP", 
and  Oq  =  OA'-  AQ\ 
Now,   if    the   chords   AB  and    AC   are    equal,   then 


a^OMETRT. 


45 


AF'=:AQ'  Q^o.U);   and  therefore  OF' =  0Q%  smd 
OF  =  OQ. 
If  the  chords  are  unequal,  suppose  AO^AB.     Then 

AQ'> AF" ;  and  therefore  OQ<OF.    Q.  E. D. 


PEOPOSITIOl^  III. 

46.  A  line  perpendicula/r  at  the  extremity  of  a  radius 
is  tangent  to  the  circle. 

Froof.  Let  the  line  ^T'  be  perpendicular  at  the  ex- 
tremity A  of  the  radius 
OA.  Take  on  ^T  any 
point  B  and  join  OB.  This 
line  OB  will  be  the  hypo- 
tenuse of  the  right-angled 
triangle  OAB^  and  there- 
fore will  be  greater  than 
the  radius  OA;  conse- 
quently the  point  B  will 
lie  without  the  circle. 
Hence  AT  will  not  touch  the  circle  anywhere  but  at 
the  point  A.  It  is  therefore  tangent  to  the  circle. 
Q.  E.  D. 

Cor.  Conversely,  if  a  line  is  toMgent  to  the  circle^  it  is 
perpendicular  to  the  radius  drawn  to  the  point  of  con- 
tact. For  the  point  of  contact  is  the  point  where  the 
tangent  is  nearest  to  the  centre.  Hence  -4(9  is  the 
shortest  line  from  0  to  AT ;  and  the  shortest  line  from 
a  point  to  a  given  line  is  the  perpendicular  (No.  19). 


46  ELEMENTS  OF 


PEOPOSITIOIS^  lY. 

47.  Three  points^  not  in  a  straight  line^  heing  given, 
one  circumference,  and  only  one,  can  he  made  to  jpass 
through  them. 

"^Proof.  Let  A,  B,  (7  be  the  three  points.  Draw  AB 
and  BG ;  bisect  the  first  in  Jtf  and  the  second  in  N,  and 
draw  MO  and  NO  perpendicu- 
lar to  AB  and  BG  respectively. 
The  point  0  will  be  the  centre 
of  a  circle  passing  through  A, 
B,  and  G.  For  drawing  OA, 
OB,  OG,  the  triangles  AOB 
and  BOG  will  be  isosceles  (]^o. 
15,  Cor.  lY.);  hence  AO,  BO, 
and  GO  will  be  equal,  as  they 
must  be  in  order  that  the  circum- 
ference described  from  0  as  centre  may  pass  through  the 
given  points.  As  any  point  0'  distinct  from  0  would 
give  unequal  values  for  the  lines  AO' ,  BO' ,  and  GO' ,  it 
is  plain  that  no  other  circumference  can  be  made  to  pass 
through  the  given  points.     Q.  E.  D. 


PKOPOSITIOK  Y. 

48.  An  angle  inscribed  in  a  circle  is  measured  hy 
half  the  arc  ojpjposite. 

Proof.  Angles  inscribed  in  a  circle  are  angles  whose 
vertices  are  on  the  circumference  of  a  circle,  and  whose 
sides  are  two  chords.  Such  are  the  angles  BA  G,  BAD, 
BAE,  the  centre  of  the  circle  being  either  on  one  of  the 
sides,  or  between  the  two  sides,  or  entirely  without.     In 


GEOMETRY. 


47 


the  first  case,  as  in  the  angle 
BAG,  it  is  plain  that  if  we 
draw  the  radius  BO,  the  angle 
BOG  will  be  twice  as  great  as 
the  angle  BA  G ;  for  the  ex- 
terior angle  equals  the  sum  of 
the  two  opposite  interior  an- 
gles, which  in  our  case  are 
equal.  Hence,  since  the  arc 
BG\s>  the  measure  of  the  an- 
gle BOG,  one-half  of  the  same 
arc  BG\^  the  measure  of  the  angle  BA  G. 

Jn  the  second  case,  as  in  the  angle  BAD,  drawing 
OD  and  OB,  we  have  one-half  of  BG  for  the  measure 
of  the  angle  BA  G,  and  one-half  of  GD  for  the  measure 
of  the  angle  GAD.  And  since  BAD  is  the  sum  of  the 
two  angles,  its  measure  will  be  the  sum  of  their  mea- 
sures, that  is,  half  the  arc  BD. 

In  the  third  case,  as  in  the  angle  DAE,  since  DAJS=: 
E AG— DAG,  the  measure  of  DAE  will  be  the  diffe- 
rence of  the  measures  of  EA  G  and  DA  G,  that  is,  half 
the  arc  DE.     Q.  E.  D. 

Gov.  I.  All  the  angles  in- 
scribed in  the  same  segment 
of  a  circle  are  equal:  as 
BAF,  BGF,  BDF,  BEE. 
For  they  are  all  measured  by 
the  same  measure,  viz.,  by  one- 
half  of  the  arc  BE. 

Gor.  II.  All  the  angles  in- 
scribed   in  a  semicircle    are 
right  angles.     For  they  are  all 
measured  by  half  the  semi-circumference,  that  is,  by  90°, 
which  is  the  measure  of  a  right  angle  (No.  43). 


48 


ELEMENTS  OF 


PKOPOSITION  YI. 

49.  Arcs  intercepted  hy  jparallel  chords,  or  ly  a  tan- 
gent and  ajparallel  chord,  are  equal. 

Proof.  Let  AB  and  CD  be  two  parallel  chords.  Join 
AD.  The  alternate  angles  BAD  and  CDA  will  be 
equal ;  hence  their  measures 
will  be  equal.  J^ow,  BAD 
is  measured  by  half  the  arc 
BD,  and  CDA  by  half  the 
arc  A  C.  Therefore  BD  and 
A  C  are  equal. 

ISTow,  let  ET  be  a  tangent 
parallel  to  the  chord  CD. 
Since  the  radius  OE  is  per- 
pendicular to  the  tangent 
and  to  the  chord  CD,  the  arc  CED  will  be  bisected  in 
E  i^o.  44).  And  therefore  the  arcs  intercepted  bj  a 
tangent  and  a  parallel  chord  are  equal.     Q.  E.  D. 


r.X 

xn 



V 

0 

PEOPOSITION   YIL 

50.  If  two  chords  intersect  within  a  circle,  the  angle 
thus  formed  is  measured  hy  half  the  sum  of  the  includ- 
ed arcs. 

Proof.  Let  AB  and  CD  be 
two  chords  intersecting  at  E. 
Draw^i^'parallel  to  6^i>.  Then 
the  angles  FAB  and  DEB  will 
be  equal  (No.  6,  Cor.)  But  the 
measure  of  FAB  is  one-half  of 
the  arc  FB,  that  is,  one-half  of 
FD-^DB,  or  one-half  of  AC+ 
DB  {Eo.  49),  which  is  the  sum 
of  the  included  arcs.    Q.  E.  D. 


SfEOMETRY, 


49 


PKOPOSITION   yiii. 

51.  The  angle  formed  hy 
two  secants  meeting  without 
the  circle^  is  measured  hy  half 
the  difference  of  the  included 
arcs. 

Proof  Let  AB  and  CB  be 
two  secants  meeting  at  B.  Draw 
Di^  parallel  to  BC.  The  angles 
ABF  and  ABC  will  be  equal 
(]S"o.  6) ;  hence  the  measure  of 
ADF,  that  is,  one-half  of  AF, 
will  also  be  the  measure  of 
ABC.  ^ui  AF=AC-FG 
=^(7— Z>^,  which  is  the  difference  of  the  arcs  inter- 
cepted.    Hence,  etc.     Q.  E.  D. 


PEOPOSITION  IX. 

52.  The  angle  formed  hy  a  tangent  and  a  chord  is 
measured  hy  one-half  of  the  intercejpted  arc. 

Proof  Let  J.r  be  a 
tangent,  and  AB  a  chord. 
Draw  the  radius  OA^  and 
the  line  OG  perpendicu- 
lar to  the  chord.  Then 
the  angles  A  0  6/and  TAB 
will  be  equal  (No.  39). 
But  the  angle  AOG  is 
measured  by  its  corre- 
sponding arc,  that  is,  by 
one-half  of  AB.  Hence  the  angle  TAB^  too,  is  mea- 
sured by  one-half  of  AB.     Q.  E.  D. 


50 


ELEMENTS  OF 


PKOPOSITIOlSr  X. 

53.  The  angle  formed  by  a  secant  and  a  tangent  is 
measured  hy  half  the  difference  of  the  arcs  intercejpted. 

Proof.  Let  the  secant  BT 
meet  the  tangent  AT  2ii  the 
point  T.  From  (7,  where  the 
secant  cuts  the  circle,  draw  CD 
parallel  to  AT,  Then  the  an- 
gles BTA  and  BCD  will  be 
equal  (No.  6).  Now,  the  angle 
BCD  is  measured  by  one-half 
of  BD,  that  is,  by  one-half  of 
BA  —AD,  or  by  one-half  of 
BA~-AC;  for  AC=:AD 
(No.  49).  Therefore  the  angle 
BTA  also  is  measured  by  BA 
^AC.    Q.E.D. 


PKOPOSITION    XI. 

54.  When  two  chords  intersect,  the  ^product  of  ths  seg- 
ments of  the  one  equals  the  product  of  the  segments  of 
the  other. 

Proof  Let  AB  and  CD 
be  two  chords  intersecting  at 
K  Join  AC  ^ndBD.  The 
triangles  ACi:  and  BDi: 
will  be  similar ;  for  the  an- 
gles A  and  D  have  the  same 
measure,  as  also  the  angles  C 
and  B.     Hence 

AE'.EDwCE'.EB; 
therefore      AExEB=  CExED.     Q.  E.  D. 


GEOMETRY. 


51 


PKOPOSITIOJSr    XII 
55.  Two  secants  meeting  at 
a  jpoint  without  a  circle  are  to 
each  other  inversely  as  their  ex- 
terior segments. 

Proof.  Let  AG  ?c^d.  BC  he 
two  secants  meeting  at  C,  and 
DC,  EG  their  exterior  seg- 
ments. Join  AE  and  BD. 
The  triangles  AGE  and  BGJ)  \ 
will  be  similar;  for  their  an- 
gles have  respectively  the  same 
measures.  Hence  the  propor- 
tion 

AG  \BG\\GE  :  GD 


Q.  E.  D. 


PKOPOSITION  XIII. 

56.  7/^  a  secant  and  a  tangent  meet  at  a  point,  the 
tangent  is  a  mean  proportional  hetween  the  secant  and 
its  exterior  segment. 

Proof.  Let  the  tangent  AT 
and  the  secant  BT  meet  at  T, 
and  let  GT  be  the  exterior  seg- 
ment of  the  secant.  Join  AB 
and  A  G.  The  triangles  ABT 
and  A  GT  will  be  similar ;  for, 
besides  a  common  angle  T,  they 
have  GBAz=:z  GAT  (No.  52); 
and  consequently  the  third  an- 
gles BAT  and  A  GT  y^iW  also 
be  equal.     Hence 

BT\AT\\AT\  GT.    Q.E.D. 


52 


ELEMENTS  OF 


PKOPOSITIOJSr  XIY. 

57.  The  side  of  a  regular  hexagon  inscribed  in  a  cir- 
die  is  equal  to  the  radius  of  the  circle. 

Proof  Let  ABCDEFhQ  a  regular  inscribed  hexagon. 
Draw  the  radii  OA,  OB,  etc. 
Since  each  side  subtends  the 
sixth  part  of  the  circumfer- 
ence, each  angle  AOB^BOC, 
etc.,  will  be  =60°.  On  the 
other  hand  (]^o.  20),  the  an- 
gles at  the  vertices  A,  B,  O, 
etc.,  are  each  =120°  ;  hence 
0^^=60°,  OBA=:60°,  etc., 
and  all  the  triangles  are  equi- 
angular and  equilateral ;  accordingly  AB  =iAO,  etc. 
Q.  E.  D. 


PEOPOSITIOK    XY. 

5S,  The  area  of  a  regular  jpolygon  is  equal  to  half 
the  jproduct  of  its  perimeter  hy  the  aj^othem. 

Proof.  Let  ABCDEF  be  a  regular  polygon,  0  its 
centre,  OH  its  apothem,  or 
the  radius  of  the  inscribed 
circle.  Drsiw  0 A,  OB,  00, 
etc.  The  polygon  will  be 
thus  divided  into  triangles, 
whose  bases  will  be  the 
sides  of  the  polygon,  and 
whose  altitude  will  be 
equal  to  the  apothem.  But 
the  area  of  any  triangle  is 
equal  to  half  the  product 
of  its  base  by  its  altitude  (No.  23).     Therefore  the  area 


GEOMETRY.  53 


of  the  polygon,  which  is  the  sum  of  the  areas  of  such 
triangles,  is  equal  to  half  the  product  of  the  perimeter 
into  the  apothem.     Q.  E.  D. 


r^^^PKOPOsiTioisr  xyi. 

59.  The  areas  of  an  inscribed  and  a  circumscrihed 
regular  polygon  of  ^n  sides  can  he  expressed  in  terms 
of  the  areas  of  the  inscribed  and  circumscribed  regular 
polygons  of  n  sides. 

Proof.  Let  p  be  the  area  of  the  inscribed,  and  P  that 
of    the  circumscribed, 

polygon    of    n    sides  ;     B E A  D c 

and  let  p'  be  the  area 
of  the  inscribed,  and 
P'  that  of  the  circum- 
scribed, polygon  of  2?i 
sides. 

Let  OIIKhQ  one  of 
the  n  triangles  which 

compose  the  area^/  if  we  represent  the  area  of  OHKhj 
a,  we  shall  have 

na=:p. 

Let  OBG  be  one  of  the  n  triangles  which  compose 
the  area  P ;  if  we  represent  the  area  of  OBC  by  J,  we 
shall  have 

nb  =  P. 

Let  GAIT  he  one  of  the  2n  triangles  that  compose  the 
area  p\  Kepresenting  the  area  GAB  by  x,  we  shall 
have 

2nx  :=p\ 

Let  GDE  be  one  of  the  27*  triangles  which  compose 


54  ELEMENTS  OF 


the  area  P' ;  if  we  represent  the  area  ODE  by  y,  we 
shall  have 

2ny—F', 

JSTow,  wehave  OMxMH=a^  OAxAB=l;  hence 

OM.ABxOA,  MH=.  ah. 

But  the  similar  triangles  OMII  and  OAB  give 

OM'.MHwOA  \AB; 

hence  OM .AB^OA,  MH, 

Therefore  {OA  .  MH)'^  ah. 

But  OA  .  MII=.^x ;  hence  ^x^—ab^  and  consequently 


4^V=  ^«  .  nJ  =J^Py  and  2?ia?  =  y  jpP  ; 
and,  since  2/ia?  =  ^',  hence 


Such  is  the  expression  of  the  area  of  the  inscribed  poly- 
gon of  2?2/  sides  in  terms  of  the  areas  of  the  polygons  of 
n  sides. 

And  now,  since  OE  bisects  the  angle  A  OB^  we  have 
(No.  37) 

AE:EB::OA:  OB, 

But  OA:  OB::  OM:  OH::OM:  OA;  and  therefore 

AE:EB::OM:  OA, 

hence       AE :  AE-\-EB  : :  OM :  0M-\-  OA, 

,      .  ^E  _       OM 

that  IS,  AB-  OM-^OA: 

Multiply  and  divide  the  first  member  of  this  equation 
by  OA,  and  the  second  member  by  MH,  We  shall 
then  have 


GEOMETRY.  55 


AExOA  OMxMH 


ABxOA  ~  OMxMH-^OAxMR 

But  AE.OA  =  y,    AB  .  0A=  h, 

OM.  MH=  a,     OA  .  MH^^x; 

-,       i>  y  a 

therefore  ^  =  — r'lr  ' 

.  2ny  na 

hence  pr-z  = r~^ — 

2no       na  -f-  2nx 

But  2ny  =  P\  2nb  ==  2P,  na  =p,  2nx  =  p\    Therefore 

Such  is  the  expression  of  the  area  of  the  circumscribed 
polygon  of  2?i  sides  in  terms  of  the  areas  of  the  poly- 
gons of  n  sides.     Q.  E.  D. 


PEOPOSITIOlSr  XYIL 

60,  The  approximate  value  of  the  area  of  a  circle^ 
whose  radius  =1,  is  3.14159265. 

Proof.  The  circle  may  be  considered  as  a  regular  poly- 
gon having  an  innumerable  multitude  of  sides  of  an  ex- 
tremely small  dimension.*     For  both  the  inscribed  and 

♦  It  has  often  been  objected  that  a  curve  cannot  be  made  up  of  straight  lines. 
But  this  stale  objection  is  easily  disposed  of.  To  say  that  a  curve  consists  of  straight 
lines  oi.  finite  length  is  certainly  absurd  ;  but  to  say  that  a  curve  consists  of  straight 
infinitesimal  elements  is  to  state  a  clear  and  necessary  truth,  about  which  mathe 
maticians  could  never  have  hesitated,  had  they  kept  in  view  the  fact,  so  clearly 
pointed  out  by  Sir  Isaac  Newton,  that  all  lines  are  traced  by  continuous  movement. 
An  infinitesimal  line  is  traced  by  an  infinitesimal  movement,  that  is,  by  the  linking 
of  two  consecutive  points.  Now,  between  two  consecutive  points  there  is  no  room 
for  a  third  point,  without  which  no  curvature  can  be  conceived.  Hence  no  infinitesi- 
mal line  can  be  a  curve.  And  therefore  all  the  infinitesimal  elements  of  curves  are 
essentially  rectilinear.  It  is  therefore  perfectly  rational  to  consider  the  circle  as  a 
regular  polygon  having  an  infinite  multitude  of  InflniteBimal  sides. 


56 


ELEMENTS  OF 


the  circumscribed  polygons,  by  doubling  again  and  again 
the  number  of  their  sides,  tend,  the  one  by  increasing, 
the  other  by  diminishing,  to  assume  equal  dimensions, 
so  that  the  difference  of  their  areas  can  be  reduced  to  an 
exceedingly  small  quantity,  and  the  area  of  the  one  can 
be  considered  practically  equal  to  the  area  of  the  other. 
A  fortiori  either  of  them  can  be  taken  as  the  area  of  the 
circle  which  is  intermediate  between  them. 

Is'ow,  starting  from  two  squares,  the  one  inscribed 
within,  and  the  other  circumscribed  about,  a  circle  whose 
radius  is  unity,  the  formulas  above  found  (]S'o.  59)  give 
us  the  following  areas  : 


Number  of 
Sides. 

Inscribed  Polygons. 

Circumscribed 
Polygons. 

4 

2.0000000 

4.0000000 

8 

2.8284271 

3.3137085 

16 

3.0614674 

3.1825979 

32 

3.1214451 

3.1517249 

64 

3.1365485 

3.1441184 

128 

3.1403311 

3.1422236 

256 

3.1412772 

3.1417504 

512 

3.1415138 

3.1416321 

1024 

3.1415729 

3.1416025 

2048 

3.1415877 

3.1415951 

4096 

3.1415914 

3.1415933 

8192 

3.1415923 

3.1415928 

16384 

3.1415925 

3.1415927 

From  this  table,  which  might  be  further  extended,  it 
is  obvious  that  the  area  of  the  circle  whose  radius  =  1, 
will  be  very  approximately  expressed  by  the  number 
3.14159265.  The  exact  area  cannot  be  expressed  in 
finite  terms,  as  its  calculation  involves  a  process  in  in- 


/J' 


GEOMETRY.  57 


fiiiitum ;  but  it  can  be  approached  nearer  and  nearer 
indefinitely.  It  is  usually  designated  by  the  Greek  let- 
ter n^  whose  value  has  been  found  by  the  infinitesimal 
calculus  to  be 

Tz  =  3.1415,926535897932384626433832795  .  .  . 

Oor.  The  area  of  a  circle  whose  radius  is  r  will  be 
7rr\  For,  as  all  circles  are  similar,  their  areas  are  to  one 
another  as  the  squares  of  the  radii,  which  are  homolo- 
gous lines. 


PEOPOSITIOISr  XYIII. 

61.  The  circumference  of  a  circle,  whose  radius  =z  1, 
is  27r. 

Proof  Let  dbcdmn  be  a  circle.  Draw  the  radii  Oa, 
Oh,  Oc,  Od,  .  .  .  ,  dividing  the  circle  into  a  great  num- 
ber of  very  small  equal  sectors.  The  sum  of  all  these 
sectors  will  be  the  area  tc  of 
the  circle.  Now,  if  the  arcs 
ab,  he,  cd,  .  .  .  are  exceed- 
ingly small  as  compared  with 
the  radius,  the  area  of  each  ' 
sector  will  be  expressed,  like 
the  area  of  a  triangle,  by  half 
the  product  of  its  base  into  its 
altitude,  the  base  being  the 
arc,  and  the  altitude  the  radius.  Hence  the  area  of  the 
circle  whose  radius  is  =1  will  be  the  sum  of  the  areas 
^ah,  ihc,  \cd,  ...  or  \{ah-\-hG-\-cd-{-  .  .  .  ).  But 
the  sum  of  all  the  small  arcs  is  equal  to  the  whole  cir- 
cumference. Therefore,  designating  the  circumference 
by  (7,  we  have  ;r=|(7/  and  consequently  (7=2;r. 
Q.E.D. 


58  ELEMENTS  OF 


Cor.  1.  The  circumference  of  a  circle  whose  radius  is 
r  will  be  'ircr.  For,  as  all  circles  are  similar,  their  linear 
dimensions  are  to  one  another  as  the  radii,  which  are  ho- 
mologous lines. 

Cor.  II.  The  number  n  expresses  the  length  of  a  semi- 
circumference  having  the  radius  =1.     Hence  -^  ex- 

presses  the  length  of  the  arc  of  one  degree  in  that  cir- 
cumference.    Thus,  when  /"  =  1,  we  have 

1°=  0.017453292, 

also  l'r=  0.000290888, 

and  r=:  0.00000484:8. 


GEOMETRY. 


59 


BOOK    lY 


GEOMETRIC  CONSTRUCTIONS. 

62.  As  Geometry  cannot  be  understood  without  dia- 
grams, every  student  of  Geometry  should  know  how  to 
construct  all  the  needed  figures,  and  how  to  ascertain 
that  they  are  accurately  constructed.  To  this  end  we 
give  here  the  solution  of  a  number  of  problems  on  geo- 
metric constructions,  as  a  practical  application  of  the 
theorems  above  established. 


PEOBLEM  I. 

63.  To  Insect  a  given  straight 
line. 

Solution. — Let  AB  be  the 
given  line.  From  A  and  B,  as 
centres,  with  a  radius  greater  A.^ 
than  one-half  of  AB  describe 
arcs  intersecting  at  E  and  F, 
and  draw  EF.  This  line  will 
bisect  AB  at  the  point  C  (]S"o. 
15,  Cor.  I.) 


% 


PEOBLEM  II. 

64.  To  erect  a  perpendicular  to  a  line  at  a  given, 
point. 

Solution. — Let  AB  be  the  line,  and  C  the  given  point. 


60 


ELEMENTS  OF 


^ 


Lay  off  the  equal  dis- 
tances CD  and  CE^  and 
from  D  and  £J,  as  cen- 
tres, with  a  radius  great- 
er than  one-half  of  DE    ^ g- 

describe   arcs   intersect- 
ing at  E,  and  draw  EC.     This  line  will  be  perpendicular 
to  AB  at  the  point  0  Q^o.  14,  Cor.) 


PKOBLEM  III. 

65.  To  erect  a  perpendicular  at  the  end  of  a  given 
line. 

Solution. — Let    AB 
be  the  given  line.  Take 
any   point    0    without 
the  line,  and  from  0, 
as  centre,  with  a  radius 
OB   describe   a    semi- 
circle,  and    from    the 
point  C  where  the  line 
AB  is  intersected  draw 
the  diameter  CD.     Then  draw  DB. 
perpendicular  to  ^^  at  its 
extremity  B  (]S:  o.  48,  Cor. 
IL) 

Or  else,  measure  three 
lengths  from  B  to  any 
point  (7  on  the  given  line, 
then  from  (7,  as  centre, 
with  a  radius  of  ^yq 
lengths,  and  from  B^  as 
centre,  with  a  radius  of 
four  lengths,  describe  arcs 


This  line  will  be 


GEOMETRY. 


61 


intersecting  at  D.  Draw  DB.  This  line  will  be  per- 
pendicular to  tlie  line  AB  at  its  extremity  B.  For 
since  CD'—  25,  GB'—  9,  and  BD'z^  16,  and  25  =  16+9, 
(7Z>  will  be  tlie  hypotenuse  of  a  right-angled  triangle 
(No.  27). 


PKOELEM  lY. 

66.  i^^m  ^  point  without  a  line  to  draw  a  perpen- 
dicular to  the  line. 

S'olution. — Let  BD  be 
the  line,  and  A  the  given 
point.  From  ^,  as  a  cen- 
tre, describe  an  arc  cutting 
BD  in  two  points  ^  and 
F.  Bisect  the  chord  EF 
at  (7,  and  draw  A  C,  which 
will  be  the  perpendicular 
required  (No.  14,  Cor.) 


\E 


F/ 


PROBLEM  Y. 

67.  To  construct  on  a  line  an  angle  equal  to  a  given 
angle. 

Solution. — Let  CAB  be  the  given  angle,  arid  MN  2, 
given  line.  From  A^ 
as  a  centre,  with  any 
radius  AB  describe 
the  arc  BC.  Then, 
with  the  same  radius, 
f.rom  Jf  as  a  centre 
describe  an  arc  PO. 
Finally  from  P,  as  a  centre,  with  a  radius  equal  to  the 


62  ELEMENTS  OF 


chord  JBG,  draw  an  arc  cutting  PO  in  Q,  and  draw 
MQ.  The  angle  FMQ  will  be  the  angle  required 
(]^o.  17). 


PKOBLEM  YI. 
68.  To  bisect  a  given  arc^  or  a  given  angle. 

Solution. — Let  ACB  be  the 
given  arc,  and  0  its  centre.  Draw 
the  chord  AB^  and  bisect  it  by 
drawing  OG,  The  arc  AB  will 
also  be  bisected  at  C  (N^o.  44). 

To  bisect  a  given  angle  A  OB^ 
from  0^  as  a  centre,  describe  an 
arc  AB^  and  bisect  it.     The  angle,  too,  will  be  bisected. 


PEOBLEM  YII. 

69.  Irom  a  given  point  to  draw  a  line  parallel  to  a 
given  line. 

Solution. — Let  C  be  the  given  point.    From  any  point 
A  of  the  given  line  AB, 

with  J.  (7  as  radius,  draw       pf  ^ 

an  arc  CB.  Then  from  C, 
as  a  centre,  with  the  same 
radius  draw  an  arc  AD, 
andmake^Z>=:^(7.  Draw 
CD.    This  will  be  the  required  Kne  (Xo.  6,  Cor.) 


PKOBLEM  Yin. 
•70.  To  find  the  centre  of  a  given  a/rc  of  circle. 


GEOMETRY. 


63 


Solution. — Let  A  CB  be 
the  given  arc.  Draw  two 
chords,  AB  and  AC^  and 
erect  perpendiculars  at  their 
middle  points  D  and  E. 
The  point  0^  where  the 
perpendiculars  intersect, 
will  be  the  centre  of  the 
arc^6^^(Ko.  44). 


PEOBLEM  IX. 

11.  From  a  given  jpoint  to  draw  a  tcmgent  to  a  given 
circle. 

Solution. — Let  A  be  the  given  point,  and  0  the  cen- 
tre of  the  given  circle. 
Join  A  0,  and  bisect  it  '"" 

in  C.  From  (7,  as  a 
centre,  with  the  radius 
CO^  describe  a  circle 
OB  AD.  This  circle 
will  intersect  the  given 
circle  at  the  two  points 
B  and  D.  Draw  AB 
and  AD.  Both  these 
lines  will  be  tangents 
to  the  given  circle. 
For,  joining  DO  and 
BO,  the  angles  ADO 
and  ABO  will  be  right, 
as  being  both  inscribed  in  a  semi-circumference  (ISTo.  48, 
Cor.  IL)  Hence  AB  and  AD  will  be  perpendicular  to 
the  radii  OB  and  OD  respectively ;  and  therefore  they 
are  tangents  to  the  circle  (^N^o.  46). 


64 


ELEMENTS  OF 


PKOBLEM  X. 

73.  To  inscribe  a  circle  in  a  given  triangle. 

Solution. — Let  ABC  be  the  given  triangle.  Bisect 
the  angles  A  and  B  by  the  lines  AO  and  BO  meeting 
at  the  point  0  y  and 
from  0  let  fall  the  per- 
pendiculars OD,  OE^ 
PF  on  the  three  sides 
of  the  triangle.  These 
perpendiculars  will  all  be 
equal.  For  the  triangles 
A  OE  and  A  OF,  having 
a  common  hypotenuse 
A  0  and  an  equal  angle 
at  A,  are  equal  (No.  27, 
Cor.) ;  hence  0E=  OF; 
and  similarly  the  trian- 
gles BOF  and  DOB, 
having  a  common  hypo- 
tenuse OB  and  an  equal 
angle  at  B,  are  equal ; 
whence  OF^=^OD.  Con- 
sequently the  point  0  is 
equally  distant  from  the 

three  sides  of  the  triangle ;  and  if,  from  0  as  centre,  with 
the  radius  OD  we  describe  a  circle,  this  circle  will  be  the 
inscribed  circle  required. 


PKOBLEM  XL 

73.  To  divide  a  line  into  jparts  jprojportional  to  other 
given  lines. 


GEOMETRY, 


65 


Solution. — Let  A,  B^  and  C  be  three  given  lines,  and 
MN  the  line  to  be  di- 


vided. From  the  ex- 
tremity M  draw  a  line 
MS  making  an  acute 
angle  with  MN^  and 
take  upon  it  MLz=zA, 


Then 


jom 


IN. 


draw    HQ ,  and 
parallel  to  IN. 


and 
LP 

The 


parts  J/P,  PQ^  QN  oi  the  given  line  JfiT  will  be  pro- 
portional to  the  given  lines  A,  B,  C  (N'o.  32). 


PKOBLEM  XII. 

74.  To  construct  a  fourth  projportional  to  three  given 
lines. 

Solution. — Let  A,  B, 
and  C  be  the  given  lines. 
Draw  two  lines  making 
an  angle  at  J/,  take 
MZ=zA,  ZK^Con  one 
line,  and  MNz=zB  on  the 
other  line.  Join  IN,  and 
from  ^draw  ICO  parallel 
to  IN;  NO  will  be  the  line  required  (No.  32). 


PROBLEM  XIII. 

75.  To  construct  a  7nean  proportional  hetween  two 
given  lines. 

Solution. — Let  A  and  B  be  the  given  lines.     Draw 


66 


ELEMENTS  OF 


ML  =^A-\-B,  and  take 
LN^A,MNz=.B.    On 

ML  as  a  diameter  de- 
scribe a  semicircle  MOL. 
From  the  point  iV^  erect 
NO  perpendicular  to 
ML.  This  perpendicu- 
lar will  be  the  mean  pro-  A 

portional  required.    For,  g 

the  angle  MOL  being  a 

right  angle  (No.  48,  Cor.  II.),  ONi^  a  mean  proportional 
between  the  segments  of  the  hypotenuse  ML  (No.  39, 
Cor.  II.) 


PKOBLEM   XIY. 

76,  Through  a  point  within  an  angle^  to  draw  a  line 
intersecting  the  sides  of  the  angle  at  equal  distances  from 
the  given  point. 

Solution. — Let  0  be  the  given  point,  and  BAG  the 
given  angle.  Draw  OD  parallel 
to  the  side  AC oi  the  angle,  and 
on  the  other  side  take  DE=i 
AD.  From  ^draw  a  line  EF 
passing  through  the  point  0. 
The  points  E  and  F  will  be 
equally  distant  from  the  given 
point ;  for  OF  and  OF  are  to 
each  other  in  the  same  ratio  as 
ED  and  DA.,  which  are  equal 
by  construction  (No.  32). 


GEOMETRY. 


67 


PEOBLEM  XY. 

77.  To  divide  a  line  in  extreme  and  mean  ratio. 

Solution. — Let  AB  be  the  given  line.  To  divide  it 
in  extreme  and  mean  ratio  is  to  divide  it  into  two  parts, 
01  which  the  greater  shall  be  a  mean  proportional  be- 
tween the  whole  line  and  the  remaining  smaller  part. 

At  the  extremity  B  of  the 
given  line  AB  draw  BO  per- 
pendicular to  AB^  and  make  it 
equal  to  one-half  of  AB.  From 
(?  as  a  centre,  and  with  OB  as 
radius,  describe  a  circumference 
BBC,  and  draw  a  secant  AG 
passing  through  the  centre  0. 
From  ^,  as  a  centre,  with  a  ra- 
dius equal  to  the  exterior  part  [ 
AD  of  the  secant,  describe  an  ' 
arc  BE  intersecting  the  given 
line  at  E.  Then  AE  and  EB 
will  be  the  parts  required.  For 
(Ko.  56)  we  shall  have 

CA  :AB::AB: 
whence       CA  -AB  :  ABwAB- 
or,  because  ABz=z  CD  by  construction, 
CA-CD:AB::AB-AD 
that  is,  AE'.ABwEB  :  AE. 


AD, 
-AD 


AD, 
AD, 


PEOBLEM   XYI. 
78.  To  inscribe  an  equilateral  triangle  in  a  given 


circle. 


Solution. — Let  ABCDEF  be  the  given  circle,  and  0 


68 


ELEMENTS  OF 


its  centre.  Draw  the  chords 
A£,  BG,  CD,  DE,  EF,  FA, 
each  equal  to  the  radius  OA. 
The  polygon  thus  formed  will 
be  a  regular  hexagon  (No.  57). 
^oin  BD,  BE, '<mdiFB.  The 
triangle  j5i>i^  will  be  the  tri- 
angle sought. 

The  side  BE  is  perpendicu- 
lar to  the  radius  OA,  by  which 
it  is  bisected  at  /  (No.  21,  Cor.  lY.) 

Sr" 

OA=r,  we  have  BE^^r"—  ir'z^-j-;  hence 


Now,  making 


BI^ 


2  V^ ;  therefore  BE=  rVS. 


PKOBLEM  XYII. 
*79.  To  inscribe  a  regular  octagon  in  a  given  circle. 
Solution. — Let  AB  aiid  CD  be  two  diameters  of  the 

given   circle,   perpen- 
dicular to  each  other. 

The  chords  ^(7,  CD, 

DB,    and    BC    will 

form  a  square.   Bisect 

these  chords,  and  their 

arcs,  by  drawing  OF, 

OF,    OG,    and    OH, 

and  draw  the  chords 

CE,  FA,    AF,  .  ,  . 

These  chords  will  be 

the  sides  of  the  regu- 
lar octagon  inscribed 

in  the  circle. 


GEOMETRY, 


69 


The  side  AC  oi  the  square  ADJBOis  equal  to  r  V~2. 
As  to  the  side  of  the  octagon,  we  have 

j:e'z=  ap-\-ep^  aI'+  {oe-  oiy ; 


and  because  u4./=:J-46^ —     ^ 
fore 


,   and  OI=^AI,  there- 


AE 
hence 


.=-  +  (,_^^)-=.(|+>_v-,+|). 


AE'=  r'  (2  -  \^\  and  AE=  r  |/2- V2. 


PEOBLEM    XYIII. 
80.  To  inscribe  a  regular  decagon  in  a  given  circle, 

/Solution. — Let  OA  be  the  radius  of  the  given  circle. 
Divide  it  in  extreme  and 
mean  ratio  (No.  77),  and  let 
00  he  the  part  which  is  a 
mean  proportional  between 
OA  and  A  0.  Draw  a  chord 
AB  ^=^00 ;  such  a  chord 
will  be  the  side  of  the  in- 
scribed regular  decagon. 

For,  since  AB  ^:^  00,  the 
proportion  OA  :  00  ::0C :  ^^ becomes 

OA  :AB::AB:A0, 

which  shows  that  the  triangles  OAB  and  BA  0  are  simi- 
lar; and  therefore  the  latter,  as  well  as  the  former,  is 
isosceles.      Hence  AB^BG^  and  consequently  also 


70  ELEMENTS  OF 


BG^:^OC^  and  the  triangle  OCB  is  also  isosceles.  We 
have,  then,  on  the  one  hand,  the  angle  COB  equal  to 
the  angle  CBO,  and,  on  the  other,  the  angle  COB  equal 
to  the  angle  CBA  j  hence  the  angle  OB  A  is  equal  to 
the  double  of  the  angle  A  OB.  The  sum  of  the  angles 
of  the  triangle  A  OB  is  therefore 

A0B-\-^A0B-^2A0Bz=im''\ 

hence  ^  (9^  ===  :^  =  36° ; 

and  as  36°  is  the  tenth  part  of  the  circumference,  the 
chord  AB  will  be  the  side  of  a  regular  inscribed  poly- 
gon of  ten  sides. 

The  side  AB  is  determined  bj  the  proportion 

r:ABr:AB:r-  AB, 

from  which  we  obtain  AB''=^  r"—  r  .  AB  ;  hence 

V6-r 


AB 


GEOMETRY.  71 


BOOK   Y 


PLANES,  AND  POLYHEDRAL  ANGLES, 

81.  A  plane  is  a  surface  on  which  straight  lines  can 
be  drawn  in  all  directions.  It  is  evident  that  a  straight 
line  will  lie  wholly  in  a  plane,  whenever  two  of  its  points 
are  in  the  plane. 

When  two  planes  are  parallel,  they  can  never  meet ; 
but  when  they  are  inclined  towards  each  other,  they 
meet  in  their  common  intersection,  and  form  a  dihedral 
angle  (that  is,  an  angle  made  by  two  faces),  which  is 
measured  by  the  plane  angle  formed  by  two  lines  drawn 
in  the  two  planes  perpendicular  to  their  intersection. 
The  intersection  is  called  the  edge  of  the  angle.  If 
three  or  more  planes  intersect  one  another,  they  will 
form  a  trihedral  or  Oi  polyhedral  angle,  whose  faces  will 
be  the  portions  of  the  planes  lying  between  the  edges. 
The  point  where  the  edges  meet  is  called  the  vertex  of 
the  angle. 


PEOPOSITION  I. 

82.  Three  points,  not  in  a  straight  line,  determine 
the  position  of  a  plane  in  space. 

Proof.  Let  J.,  ^,  and  C" be  three 
given  points.  Join  AB,  and  con- 
ceive a  plane  passing  through,  and 
revolving  around,  AB.  When  the 
plane  reaches  the  point  C,  it  has  a 


72 


ELEMENTS  OF 


definite  jDosition ;  and  tlnis  the  three  points  determine 
the  position  of  the  plane.     Q.  E.  D. 

Cor.  I.  Two  straight  lines^  which  intersect,  determine 
the  jposition  of  a  plane. 

Cor.  11.  Two  parallel  lines  determine  the  position 
of  a  plane. 


PEOPOSITIO]^  11. 

83.  The  intersection  of  two  planes  is  a  straight  line. 

Proof  Let  AB  and  CD  be  two  planes,  and  E  and  F 
two  points  common  to  both  planes.  Draw 
the  straight  line  EF.  Because  this  line 
has  two  points  in  the  plane  AB,  the  whole 
line  will  be  in  that  plane ;  and  because  it 
has^two  points  in  the  plane  CD,  it  will  lie 
wholly  in  this  plane.  ]N^ow,  the  two  planes 
have  no  common  line  but  their  intersection. 
Hence  the  straight  line  EF,  which  is  in 
both  planes,  is  the  intersection  of  the  two  planes 
Q.  E.  D. 


PEOPOSITIOlSr  III. 

84.  J.  straight  line  perpendicular  to  two  straight 
lines  at  their  point  of  intersection,  is  perpendicular  to 
the  plane  of  those  lines. 

Proof  Let  BC  and  DE  be  two 
lines  intersecting  at  P,  and  let 
AP  be  perpendicular  to  them. 
Through  P  draw  any  line  P^  in 
the  plane  of  PE  and  PC,  and 
through  the  point  Q  draw  a  line 
EC  so  that  QC  and  QE  shall  be 
ecjual  (Ko.  76).    And  now  draw 


GEOMETRY.  73 


CA,  EA,  and  QA.  The  base  EC  of  the  triangle  AEG 
being  bisected  by  A  Q,  we  shall  have  Q^o.  30) 

AC'+  AE'^  2J^^+  ^CQ\ 

In  like  manner,  the  base  EG  of  the  triangle  PEG  being 
bisected  by  PQ-,  we  shall  have  also 

PG'-\-PE'=  2P^^+  267^'. 

Subtracting  this  from  the  preceding  equation,  we  have 

J^«_  FC'+  AE'-  PE'=  2AQ'-  2PQ\ 

But  AG'-  PG'=z  AP%  and  AE'-  PE'=  AP' ; 

hence  2AP':=  2AQ'-  2PQ'  ; 

whence  AQ'=  AP'+  PQ'  ; 

which  shows  (No.  2Y)  that  the  triangle  APQ  is  right- 
angled  at  P,  and  therefore  AP  is  perpendicular  to  ^^/ 
and  since  i^^  is  any  line  drawn  through  P  in  the  plane 
of  J5(7and  DE,  it  follows  that  AP,  if  perpendicular  to 
these  two  lines,  is  perpendicular  to  every  other  line 
drawn  through  P  in  their  plane,  viz.,  AP  is  perpen- 
dicular to  the  plane  PEG.     Q.  E.  D. 


PEOPOSITIOlSr  lY. 

S^,  The  perpendicula/r  is  the  shortest  distance  from 
a  jpoint  to  a  plane. 

Proof.  From  the  point 
A  let  fall  AP  perpendicu- 
lar to  the  plane  JO^.  Then 
all  the  oblique  lines,  as  A  G^ 
AD,  AE,  drawn  from  A  to 
the  same  plane  will  be  great- 
er than  AP ;  for  they  will 


74 


ELEMENTS  OF 


be  hypotenuses  of  right-angled  triangles,  of  which  AP 
is  a  side.  Hence  AP  is  the  shortest  line  from  A  to  the 
plane.     Q.  E.  D. 

Cor.  I.  Oblique  lines  equally  inclined,  or  equally  di- 
verging from  the  perpendicular,  are  equal.  For  they 
are  the  hypotenuses  of  equal  triangles. 

Cor.  II.  The  oblique  lines  AD  and  A  C,  meeting  the 
plane  at  equal  distances  from  the  foot  of  the  perpendi- 
cular, are  equal. 

Cor.  III.  From  a  given  point  to  a  given  plane  only 
one  perpendicular  cam,  be  drawn. 


PEOPOSITIOK  Y. 

86.  If  from  the  foot  of  a  perpendicular  to  a  plame 
a  line  be  drawn  meeting  at  right  angles  a  straight  line 
of  that  plane,  a  line  joining  the  point  of  their  in- 
tersection with  any  foint  of  the  perpendicular  will  be 
perpendicular  to  the  line  of  the  plane. 

Proof  Let  AP  be  perpendicular  to  the  plane  JfiT, 
and  BC  the  given  line 
at  right  angles  with 
PQ.  Talie  QC=QB, 
and  draw  BP,  CP, 
BA,  CA,  and  AQ. 
The  triangles  PBQ 
and  PCQ  will  be 
equal,  for  they  have 
equal  sides  about  a 
right     angle ;      hence 

PC^PB.     The  triangles  APB  and  APC  will  also 
be  equal,  for  they,  too,  have  equal  sides  about  a  right 


GEOMETRY. 


15 


angle ;  hence  A£  =  AC.  Consequently  the  triangle 
ABC  is  isosceles,  and  the  line  AQ  which  bisects  its 
base  BC  is  perpendicular  to  BO  (ISTo.  14,  Cor.) 
Q.  E.  D. 

Cor.  The  line  BCis  perpendicular  to  the  plane  of  the 
triangle  APQ  ;  for  it  is  perpendicular  to  the  two  lines 
P^  and  ^(2  (No.  84). 


PKOPOSITIOK  YL 

87.  If  one  of  two  jparallels  is  perpendicular  to  a 
plane,  the  other  also  is  perpendicular  to  that  plane. 

Proof.  Let  the  line  DQhQ  parallel  to  the  perpendicu- 
lar AP.  If  we  pass  a  plane  through  these  Hnes,  its  in- 
tersection with  the  plane  J/TT 
will  be  PQ.  Through  the 
point  Q  draw  in  the  plane 
MN  iho,  line  BC  perpendicu- 
lar to  P Q,  and  join  A  Q.  Then 
BQ^  being  perpendicular  to 
PQ  and  to  AQ  (No.  86),  will 
be  perpendicular  to  the  plane 
APQD  (No.  84);  and  there- 
fore BQD  is  a  right  angle. 
Now,  DQP  is  also  a  right  angle  ;  for  AP  and  DQ  are 
parallel.  Therefore  DQ  \s>  perpendicular  to  the  two 
lines  PQ  and  QB  lying  in  the  plane  MN ;  and  conse- 
quently (No.  84)  it  is  perpendicular  to  the  same  plane. 
Q.  E.  D. 

Cor.  I.  If  two  lines  are  perpendicular  to  the  same 
plane.,  they  are  parallel.  For,  \i  DQ  were  not  parallel 
to  AP,  we  could  draw  through  Q  another  line  parallel 
to  AP,  which  would  be  perpendicular  to  the  plane  MN^ 


^6  ELEMENTS  OF 


and  then  there  would  be  two  perpendiculars  at  the  same 
point,  which  is  impossible. 

Cor.  II.  If  two  straight  lines  are  parallel  to  a  third^ 
they  will  he  parallel  to  each  other.  For  a  plane  perpen- 
dicular to  one  of  them  will  be  perpendicular  to  the  other 
two. 


PEOPOSITIOJSr  YII. 

88.  7f  t^  straight  line  is  parallel  to  any  line  of  a 
pla/ne,  it  is  parallel  to  the  plane. 

Proof.  Let  AB  be  parallel  to  the  line  CD  lying  in 
the  plane  MN.     The  lines 

AB  and  CD,  being  paral-  A b 

lei,  lie  on  a  plane  ABCD, 
and  therefore  AB  cannot 
meet  J/IZV  except  in  some 
point  of  the  line  CD.  But 
this  is  impossible,  because 
AB  and  CD  are  parallel. 

Hence  AB  can  nowhere  meet  the  plane  MN ;  and  con- 
sequently this  line  is  parallel  to  the  plane.     Q.  E.  D. 


PROPOSITION  YIII. 

89.   If  two  planes  are  perpendicular  to  the  same 
straight  line,  they  are  parallel  to  each  other. 

Proof,  Two  planes  are 
parallel  which  can  never 
meet.  But  two  planes  per- 
pendicular to  the  same  line 
AB  can  never  meet ;  for,  if 
they  met  at  any  point  0, 
there  would  be  a  triangle 


GEOMETRY. 


77 


ABO  with  two  right  angles  GAB  and  DBA,   which 
is  impossible.     Q.  E.  D. 


PEOPOSITIOJS'  IX. 

90.  If  ajplane  intersects  two  j^arallel  jplcmes^  the  lines 
of  intersection  will  he  parallel. 

Proof  Let  JO^and  P^  be  two 
parallel  planes  intersected  by  a  plane 
ABDC.  If  the  intersections  AB 
and  CD  are  not  parallel,  they  will 
meet  somewhere ;  for  they  are  in  the 
same  plane.  But  if  these  lines  meet, 
then  the  parallel  planes  MN  and 
PQ  will  meet,  which  is  impossible. 
Hence  the  lines  AB  and  CD  are  parallel.     Q.  E.  D. 


PEOPOSITIOlSr  X. 

91.  If  two  lines  are  jpar allele  they  are  equally  in- 
clined to  any  jplane  passing  through  them. 

Proof.  Let  AB  and  CD  be  two  parallel  lines,  and 
MN"  a  plane  passing  through  them  at  A  and  C.     Take 
AB=  CD,  and  draw  AC 
and  BD.    The  quadrilateral 
ABDC  V7\\\  be  a  parallelo- 
gram (:N'o.  21) ;  hence  BD= 
AC.     Let  fall  ^^  and  PP 
perpendicular  to  the  plane, 
and  draw  JEF.     The  quadri- 
lateral BDEF  will  be  a  par- 
allelogram;   hence  BE=DF,  and  EF=BD  =  AC. 
Draw  now  AF  and  CF.    The  quadrilateral  AEFC  will 


78  ELEMENTS  OF 


also  be  a  parallelogram  ;  and  therefore  AE=  CF.  Hence 
the  two  triangles  ABE  and  CDF  have  equal  sides,  and 
the  angles  BAE  and  DCF^  opposite  to  the  equal  sides 
BE  and  DF,  are  equal.     Q.  E.  D. 


*  pkopositio:n^  xi. 

93.  Parallel  lines  hetween  pa/rallel  jplanes  are  equal. 

Proof,  Let  AB  and  CD  be  two 
parallel  lines  included  between  the 
parallel  planes  MN  and  PQ.  The 
plane  ABDC  oi  the  two  lines  will 
intersect  the  two  planes  in  the  lines 
A  C  and  BD^  which  will  be  parallel 
(]Sro.  90).  Hence  ABDC  will  be 
a  parallelogram  ;  and  consequently 
AB=CD.     Q.E.D. 


PEOPOSITION  XII. 

93.  Two  angles^  situated  in  different  jplanes^  whose 
sides  are  jparallel  and  similarly  directed^  are  equal  j 
and  their  jplanes  are  ^parallel. 

Proof.  Let  BAC  and  EDF  \iQ  two  angles,  whose 
sides  are  parallel  and  similarly 
directed.  Take  AB  =  DE  and 
AC=DF,  and  draw  ^(7  and 
EF.  Since  AB  and  DE  are 
equal  and  parallel,  we  may,  by 
joining  AD  and  BE,  form  a 
parallelogram  ABED  i  also, 
since  A  C  and  DF  are  equal  and 
parallel,  A  CFD  will  be  a  paral- 


u 

D               / 

/'\vl'  / 

. 

/     \           \     1_  / 

/°" 

-^ 

y 

GEOMETRY, 


79 


lelogram;  hence  BE=AD^CF.  But  if  BE=CF, 
BEFG  will  also  be  a  parallelogram  ;  and  therefore  BG 
=  EF.  Thus  the  triangles  ABC  and  DEF  will  be 
equal ;  and  accordingly  the  angles  BA  C  and  EDF  will 
be  equal.     Q.  E.  D. 

Again,  since  AB  and  DE  are  parallel,  they  cannot 
meet ;  also  A  C  and  J)F,  being  parallel,  cannot  meet ; 
hence  the  plane  determined  by  ^^  and  AG  cannot 
meet  the  plane  determined  by  BE  and  BE;  that  is, 
the  planes  of  the  angles  BA  G  and  EDF  are  parallel. 
Q.E.D. 

Gov.  Two  planes,  as  AGED  and  BGFE,  intersecting 
two  parallel  planes,  form  equal  angles,  as  A  GB^  BEE, 
by  their  intersections  with  them. 


PROPOSITIOIS'  XIII. 

94,  Straight  lines^  if  cut  by  parallel  j>lanes,  will  he 
divided  into  jprcyportional  jparts. 

Proof,  Let  any  two  straight  lines  AB  and  GB  be 
cut  by  the  parallel  planes  MN^ 
PQ,  JST.  Draw  AB.  Since 
PQ  and  /ST  are  parallel,  the 
lines  EF  and  BB,  in  which 
they  are  met  by  the  plane 
ABB,  are  parallel  (No.  90). 
Hence  the  triangles  ABB  and 
AEF  are  similar,  and  we  have 

AE:BE::AF:BF 

And  since  JfiV  and  PQ  are 
parallel,  the  lines  A  G  and  EG, 
in  which  they  are  met  by  the 


M 

/• 

n7 

P_ 

\        1 

/  = 

^r1°  / 

S 

\  1  ^ 

/' 

^'/ 

80 


ELEMENTS  OF 


plane  ADC,  are  parallel,  and  the  similar  triangles  A  CD 
and  FGD  will  give 

AF  \DF\\  CG\DG, 

Combining  these  two  proportions,  we  obtain 
AE\BE\\Ca\DG.    Q.E.D. 


PEOPOSITIOIT  XIY. 

95.  ^  a  straight  line  is  jperjpendiculaT  to  a  plane^ 
all  planes  passing  through  that  line  will  he  jperjpendi- 
cular  to  that  'plane. 

Proof,  Let  AP  be  perpendicular  to  the  plane  MN, 
and  let  a  plane  BE  pass  through  AP.  In  the  plane 
MN  draw  the  line  PD  perpendi- 
cular to  the  intersection  EC  oi  the 
two  planes ;  then,  because  ^P  is 
perpendicular  to  the  plane  MN,  the 
angle  APD  will  be  a  right  angle  ; 
and  this  angle,  because  it  is  formed 
by  two  perpendiculars  to  the  inter- 
section BC  oi  the  two  planes,  is 

the  measure  of  the  divergence  of  the  two  planes.  Hence 
the  two  planes  are  at  right  angles  ;  and  thus  every  plane 
passing  through  a  line  perpendicular  to  a  plane,  is  per- 
pendicular to  that  plane.     Q.  E.  D. 


PKOPOSITION    XY. 
96.  If  two  planes  are  perpendicular  to  each  other. 


GEOMETRY. 


81 


any  line  drawn  in  one  of  them  jperpendicular  to  their 
intersection^  will  he  jperj^endicular  to  the  other. 

Proof.  Let  the  planes  MN  and  BE  be  perpendicular 
to  each  other,  and  let  the  line 
AP^  drawn  in  the  plane  BE^ 
be  perpendicular  to  the  inter- 
section BC.  Then,  since  the 
two  planes  are  perpendicular, 
APD  will  be  a  right  angle; 
and  therefore  AP  is  perpendi- 
cular to  the  two  lines  BC  and 

PD  at  their  intersection.     It  is  therefore  perpendicular 
to  their  plane  MN,    Q.  E.  D. 


M 

i\^.^ 

E 

/. 

^^ 

1/ 

PKOPOSITION  XYI. 

97.  The  intersection  of  tioo  planes  perpendicular  to 
a  third  plane  is  perpendicular  to  that  plane. 

Proof  Let  the  planes  BE  and  BE  be  perpendicular 
to  the  plane  MN,  and  let  J.P 
be  their  intersection.  If  from 
P  we  erect  a  perpendicular  to 
the  plane  MN,  such  a  perpen- 
dicular must  be  in  the  plane 
BE,  and  also  in  the  plane  JDE 
(IS'o.  96),  and  is,  therefore, 
their  common  intersection  AP. 
Q.  E.  D. 

Cor.  A  plane  perpendicular  to  two  other  planes  is  per- 
pendicular to  their  intersection. 


82 


ELEMENTS  OF 


'  PEOPOSITIO]^  XYII. 

98.  The  sum  of  any  two  jplane  angles  formed  hy  the 
edges  of  a  trihedral  angle  is  greater  than  the  third 
angle. 

.  Proof  Let  AYB,  BVC,   and  OVA  be  the  three 
plane  angles,  and  let  A  VjB  be  the  greatest.     In   the 
plane  A  VB   make  an  angle 
BYD   equal   to   BVC,   and 
draw  AB  at  pleasure.     Take 

YC=  YD,  and  draw  A  C  and 
CB.  The  triangles  CYB 
and  B  YD  will  be  equal  (N^o. 
11) ;  and  therefore  BD  =BG. 
But  (:^ro.  13) 

AC-\-BC>AD-^DB; 

hence,  cancelling  the  equal  parts  BC  and  DB,  we  have 
AC>  AD;  and  consequently  (ISTo.  18)  AYC>A  YD; 
and,  adding  the  equal  angles  B  YC  and  B  YD, 

AYO+BYOAYD  +  BYD, 
that  is,       AYC+BYOA  YB.     Q.  E.  D. 


PEOPOSITIOIS'  XYIII. 

99.  The  sum  of  the  plane  angles  around  a  polyhedral 
angle  is  always  less  than  four  right  angles. 

Proof.  Let  T^be  the  vertex  of  any  polyhedral  angle. 
By  passing  a  plane  through  its  edges  YA,  YB,  YC,  .  .  . 


GEOMETRY.  •  83 


we  shall  form  a  set  of  triangles 
AYB,  BYC,  ,  .  ,  \  and,  if 
from  any  point  0  in  the  plane 
ABC  ...  we  draw  the  straight 
lines  OA,  OB,  00,  .  .  .  we 
shall  form  another  set  of  trian- 
gles ^6>^,  ^0(7,  .  .  . 

Let  n  be  the  number  of  the 
triangles  of  each  set.  The 
snm  of  the  angles  of  all  the  triangles  of  the  first  set 
will  be  180°  X  w,  and  will  consist  of  two  parts,  viz.,  of 
the  angles  about  F",  the  sum  of  which  we  may  designate 
by  (F),  and  of  the  angles  YBA,  YBO,  YOB,  .  .  . 
around  the  base.     We  may  then  write 

(Y)  +  {YBA  +  YBO+YOB+  .  .  .  )  =  180°x^. 

The  sum  of  the  angles  of  all  the  triangles  of  the  second 
set  is  also  180°  X  n,  and  consists  of  two  parts,  viz.,  of 
the  angles  about  0,  the  sum  of  which  we  may  designate 
by  {0),  and  of  the  angles  OBA,  OBO,  OOB,  .  .  . 
"We  may  therefore  write 

{0)  +  {OBA+OBO+OOB+  .  .  .  )  =  180°x^. 

Hence 

(Y)  +  {YBA  +  YBO+YOB+  .  .  .  ) 

=  {0)  +  {OBA  +  OBO+OOB+  .  .  .  ). 

^Now,  if  the  sum  YBA-{-YBO+  ...  is  greater  than 
the  sum  OBA-\-OBO-{-  .  .  .  ,  it  is  obvious,  from  this 
equation,  that  ( Y)  must  be  less  than  (0).  But  we  have 
proved  (No.  98)  that  YBA  +  YBO  is  greater  than 
ABO,  that  is,  YBA  +  YBO>  OBA  +  OBO.  And, 
since  the  same  is  true  also  of  the  other  angles  around  the 


84  ELEMENTS  OF 


polygon  ABC  .  .  .  ,  we  conclude  that  (  V)  is  less  than 
(6^).  But  (6^)  represents  four  right  angles.  Hence  (  V) 
is  less  than  four  right  angles.     Q.  E.  D. 


PEOPOSITIOK  XIX. 

100.  If  the  jplcme  angles  at  the  vertex  of  a  trihedral 

angle  are  equal,  each  to  each,  to  the  plane  angles  at  the 
vertex  of  another  trihedral  angle,  the  planes  of  the  equal 
angles  are  equally  inclined  to  each  other. 

Proof  Let  J[f  and  iTbe  the  vertices  of  the  trihedral 
angles,  and  let  AMC  =  DNF,  AMB=DNE,  BMC= 
ENF.  Draw  AB  and 
AC,  both  perpendicular  to 
AM.  Then  will  the  angle 
BA  C  measure  the  inclina- 
tion of  the  two  planes 
MAB2.v.^MAC  Layoff 
ND  equal  to  MA,  and 
draw  DE  and  DF  both 
perpendicular  to  ND. 
Then  will  the  angle  EDF 
measure  the  inclination  of 
the  two  planes  NDE  and 
NDF. 

Now,  the  right-angled  triangles  MAB  and  NDE 
have  the  sides  ^JtTand  DN  equal  by  construction,  and 
the  angles  AMB  and  DNE  equal  by  supposition. 
Hence  AB  =  DE  and  MB  =  NE. 

Similarly,  the  right-angled  triangles  MA  C  and  NDF 
have  the  sides  AM  and  DN  equal,  and  the  angles 
AMC  and  DNE  equal.  Therefore  AC—DF  and 
MC=zNF. 


GEOMETRY.  85 

Finally,  the  triangles  CMB  and  FMJE  have  the  equal 
angles  CMB  and  FNE  comprised  between  equal  sides. 
Hence  these  triangles  are  equal,  and  therefore  CB—FE. 
The  consequence  is,  that  the  triangles  ABC  and  DEF 
are  equal,  and  that  the  angles  CAB  and  FDE,  which 
measure  the  inclination  of  the  respective  planes,  are 
also  equal.  And  the  same  conclusion  holds  with 
regard  to  the  inclination  of  the  other  planes;  that 
is,  the  planes  of  equal  angles  are  equally  inclined. 
Q.  E.  D. 


86  ELEMENTS  OF 


BOOK   YL 


SUBFACES    AND     VOLUMES    OF    POLY- 
HEDBONS, 

101.  Polyhedrons  are  geometric  solids  bounded  by 
polygons.  These  polygons  are  cdll^di  faces  of  the  poly- 
hedron ;  the  lines  in  which  they  meet  are  called  edges; 
and  the  points  in  which  the  edges  meet,  vertices  of  the 
polyhedron. 

Begular  polyhedrons  are  those  whose  faces  are  equal 
and  equally  inclined,  as  the  regular  tetrahedrons,  hexa- 
hedrons, octahedrons,  dodecahedrons,  and  icosahedrons. 
All  the  other  polyhedrons  are  irregular,  because  they 
have  not  equal  faces  equally  inclined ;  to  this  class  be- 
long the  irregular  jpmms  smd  pyramids, 

A  prism  is  a  polyhedron,  of  which  two  faces  are  equal 
and  parallel  polygons,  called  bases,  the  other  faces  being 
parallelograms.  A  prism  whose  bases  are  parallelograms, 
is  called  a  parallelepipedon.^ 

A  pyramid  is  a  polyhedron  bounded  by  a  polygon 
called  the  base  and  by  triangles  meeting  at  a  common 
point,  called  the  vertex  of  the  pyramid. 

All  the  other  irregular  polyhedrons  may  be  considered 
as  made  up  of  prisms  and  pyramids  arranged  in  a  cer- 
tain order. 

*  Our  countrymen  write  parallelojnpedon ;  but  the  Greek  word  is  TrapoAAijA- 
firineSov,  and  I  see  no  reason  why  it  should  be  corrupted. 


GEOMETRY. 


87 


PEOPOSITION  I. 

102.  The  convex  surface  of  a  right  prism  is  equal 
to  the  perimeter  of  the  'base  multiplied  hy  its  alti- 
tude. 

Proof  Let  ABODE  —  I  hQ  a  right  prism,  viz.,  a 
prism  of  wliich  the  lateral  edges 
are  perpendicular  to  the  basis.  Its 
convex  surface  is  the  sum  of  all 
the  lateral  surfaces,  that  is,  of  the 
rectangles  AG.BH,  CI,  DK.^u^ 
EF.  ]^ow,  these  rectangles  have 
the  same  altitude,  viz.,  the  altitude 
of  the  prism.  Hence  the  sum  of 
their  surfaces  is 


{AB^BC^ CD+DE+EF)  AF,  B 

which  is  the  perimeter  of  the  base  multiplied  by  the  al- 
titude.    Q.  E.  D. 

Cor.  If  two  right  prisms  have  the  same  altitude,  their 
convex  surfaces  are  to  each  other  as  the  perimeters  of 
their  bases  ;  and  if  they  have  the  same  base,  their  con^ 
vex  surfaces  are  to  each  other  as  their  altitudes. 


PEOPOSITIOK  II. 

103.  The  opposite  faces  of  any  parallelepipedon  are 
equal,  and  their  planes  are  parallel. 

Proof.  Let  ABCP—G  be  a  parallelepipedon.      Its 


88 


ELEMENTS  OF 


bases  A  BCD  and  EFGH  are  equal  and  parallel  (]N"o. 
101).     Hence  EF^AB  =D  G 
^HG,  and  HE^ABz^^CB^ 
GF.      Consequently   (JSTo.  21, 
Cor.  II.) 


EA  =  FB^  CG^DH; 

hence   DAEH^GBFG,  and 
ABFE^DCGIL    Q.  E.  D. 

Since  ^D  and  6^^,  as  also  EA  and  ^i^^,  are  parallel, 
hence  (Ko.  93)  the  planes  AEHD  and  BFGG  are  par- 
allel. In  like  manner,  it  may  be  shown  that  the  planes 
ABFE  and  DCGH  are  parallel.     Q.  E.  D. 

Cor.  In  a  rectangular  parallel- 
epipedon^  the  square  of  a  diagonal 
equals  the  sum  of  the  squares  of  the 
three  edges  which  meet  at  the  sa/me 
vertex.  For,  let  ^^  be  a  diagonal 
of  the  parallelepiped  on,  and  AC  2^ 
diagonal  of  the  base.  Then  AD''= 
nC'  +  AG'' ;  and  because  AC'— 
AB'-\-  BG\  hence 


c 


AB'^  BC'Ar  ^^'+  ^^'=  i>(7'+  EC'-\-  BC\ 


PEOPOSITIOlSr  III. 

104.  If  a  plans  he  passed  diagonally  through  a 
parallelepipedon^  it  will  divide  it  into  two  equal  tri- 
angular prisms. 


GEOMETRY. 


89 


Proof.   Let  AG  \)q  2. 
parallelepipedon,  and 

BDHF  the  diagonal 
plane  which  divides  it 
into  two  prisms.  The  tri- 
hedral angles  of  the  two 
prisms,  as  formed  by  equal 
plane  angles,  will  be  re- 
spectively equal  (No.  100), 
the  triangular  bases  will 
be  equal  (No.  21),  and  the 
edges  also  equal.  Hence 
the  two  prisms  are  equal  in  all  their  parts. 


Q.  E.  D. 


PEOPOSITIOlSr  lY. 

105.  Two  parallelepipedons  having  a  common  lower 
hase.,  and  their  upper  bases  hetween  the  same  parallels, 
are  equal  in  volume.  ^ 

Proof.  Let  the  parallelepipedons  J-<9  and  J.JIf  have 
the  common  lower  base  ABCD,  and  their  upper  bases 
EFGH  and  KLMN  be- 
tween the  same  parallels 
FK^id^GN.  The  volume 
of  ^  6r  is  equal  to  the  whole 
volume  of  the  figure,  minus 
the  prism  AEK—N;  and 
the  volume  of  ^J^is  equal 
to  the  wliole  volume  of  the 

figure,  minus  the  prism  BLF—  G.  But  these  two  prisms 
are  equal ;  for  their  bases  AEK  and  BLF  have  AK^=. 
LB.  AE—  BF^  and  the  angle  ^J.^ equal  to  the  angle 
LBF^  and  therefore  (No.  11)  they  are  equal;   on  the 


90 


ELEMENTS  OF 


Other  hand,  the  faces  ^i^iiTTand  BCML,  also  AD  HE 
and  BCGF,  and  lastly  KEHIST  2.ndi  LFGM  are  respec- 
tively equal.  Hence  the  two  prisms  are  equal  in  all 
their  parts.  Hence  whether  from  the  volume  of  the 
whole  figure  we  subtract  the  one  of  the  prisms  or  the 
other,  the  remainder  will  be  the  same.  But  by  subtract- 
ing the  prism  AEK—N^  the  remainder  is  the  parallel- 
epipedon  AG ;  and  by  subtracting  the  prism  BLF—G, 
the  remainder  is  the  parallelepipedon  AM.  Therefore 
the  two  parallelepipedons  are  equal  in  volume.     Q.  E.  D. 


PEOPOSITIOJS"  Y. 


106.  Two  parallelepvpedons  having  a  common  base 
cmd  the  same  altitude,  are  equal  in  volume. 

Proof.  Let  the  parallelepipedons  AR  and  A  G  have 
the  common  bfese  A  BCD  and  the  same  altitude.  Be- 
cause they  have  the  same 
altitude,  their  upper  bases 
will  lie  in  the  same  plane. 
Prolong  EF  and  GH,  as 
also  PS  and  QE.  These 
prolongations  will  form 
a  parallelogram  KLMN 
equal  to  ABCD  and  to 
EFGH.  Now,  if  a  third 
parallelepipedon  be  con- 
structed  on  ABCD  with 

the  upper  base  KLMN,  its  volume  will  be  equal  to  that 
of  A  G,  because  the  lines  KF  and  NG  are  parallel,  and 
will  also  be  equal  to  that  of  AR^  because  the  lines  PN 


GEOMETRY. 


91 


and  QM  are  parallel  (No.  105).     Hence  the  volumes  of 
the  parallelepipedons  AB  and  A  G  are  equal.     Q.  E.  D. 
Cor.    Any  ohlique  jparallelejpipedon    can  he    trans- 
formed into  a  right  jparallelejpijpedon  homing  the  same 
base  and  the  same  altitude^  with  no  change  of  volume. 


PEOPOSITION  YI. 

10*7.  The  volume  of  a  rectangular  jparallelepi^edon 
is  measured  hy  the  product  of  its  three  dimensions. 

Proof  Let  AD  be  a  rectangular  parallelepipedon, 
and  a^  the  unit  of  measure 
adopted  for  volumes.  This 
unit  is  the  cube  of  a  unit 
of  length.  To  express  the 
volume  of  the  parallelepi- 
pedon it  suffices  to  state 
how  many  times  it  contains 
the  unit  of  measure.  [NTow, 
divide  the  three  dimensions 
of  the  parallelepipedon  bj 
the  unit  of  length,  and  as- 
sume AB^^pa,  BG  =  qa, 
CD  =  ra.     Then  the  unit 

of  measure  a""  will  be  contained  in  the  parallelepipedon 
pqr  times ;  for  a  row  of  p  units  laid  down  from  A  to 
B,  and  repeated  q  times  from  B  to  G,  will  form  a  first 
Isijerpqa'  on  the  base  of  the  parallelepipedon  ;  and  this 
layer  must  be  taken  r  times  in  ascending  from  G  to  D 
Q.  E.  D. 

From  what  has  been  said  above  (IS'o.  22)  it  can  be 
shown  that  this  proposition  is  true  whether  p,  q,  and  r 


/ 

/ 

/ 

D 

/ 

f 

C 

/ 

/ 

^  / 

^  / 

/ 

/  /  / 

/ 

/  /  /  / 

/ 

92 


ELEMENTS  OF 


be  entire  or  fractional,  and  commensurable  or  incommen- 
surable with  the  unit  of  measure. 


PKOPOsiTioisr  yii. 

108.  A  right  parallelepipedon  can  he  i/i^ansformed 
into  a  rectangular  parallelepipedon  of  the  same  alti- 
tude^ without  change  of  volume. 

Proof  Let  ABCD—G  be  a  right  parallelepipedon 
having  for  its  base  the  parallelo- 
gram ABCD.  Through  AE  and 
BF  pass  the  planes  A  Q  and  BP 
perpendicular  to  the  plane  AF. 
Their  intersections  with  the  other 
planes  will  determine  the  rectan- 
gular parallelepipedon  ABMN 
—P,  equal  in  volume  to  the  given 
parallelepipedon,  because  it  has 
an  equal  altitude,  and  a  base  of 
equal  area  (Ko.  106).     Q.  E.  D. 

Cor.  I.  Since  any  oblique  parallelepipedon  can  be 
transformed  into  a  right  parallelepipedon  having  the 
same  base,  the  same  altitude,  and  the  same  volume,  and 
since  any  right  parallelepipedon  can  be  transformed  into 
a  rectangular  parallelepipedon  having  an  equal  base,  an 
equal  altitude,  and  an  equal  volume,  it  follows  that  the 
vohime  of  a/ny  jparallelejpijpedon  can  he  expressed  hy  the 
jproduct  of  the  three  dimensions  of  a  rectangidar  paral- 
lelepipedon having  the  same  altitude  and  an  equal  hase. 

Cor.  II.  Parallelepipedons  having  equal  hases  are  to 
each  other  as  their  altitudes;  and  parallelepipedons 
having  equal  altitudes  are  to  each  other  as  their  hases. 


c 

\    H 

3 

\^ 

> 

F 
M 

y 

N 

/ 

y 

D 

• 

y 

GEOMETRY, 


ds 


For,  let  P  and  P'  be  two  parallelepipedons,  having  the 
bases  B  and  B'^  and  the  altitudes  A  and  A'  respectively. 
Then  P=  A .  B,  P'^  A',  B' ;  whence 

P\P'\\  A.B:A\B\ 

If  B=:B\  this  proportion  becomes  P  :  P' ::  A  :  A\ 
If,  on  the  contrary,  A=A',  the  proportion  becomes 
P.P'v,  B\B\ 

Cor.  III.  Triangular  prisms  are  semi-parallelepipedons 
(No.  104),  and  therefore  their  volume  is  measured  hy 
the  product  of  their  triangular  hase  (which  is  one-half 
of  the  base  of  the  parallelepipedon)  into  their  altitude. 
If  they  have  equal  bases,  they  are  to  each  other  as  their 
altitudes  /  and  if  they  have  equal  altitudes,  they  are  to 
each  other  as  their  bases. 


pkopositio:n"  yiii. 

109.  The  volume  of  any  prism  is  equal  to  the  pro- 
duct of  its  base  by  its  altitude. 


Proof.  Let  ABCDEF  —  L  be  any  prism 
a  lateral  edge  BH  draw  the  planes 
BN,  BM,  BL  dividing  the  prism 
into  triangular  prisms.  These  prisms 
will  have  the  same  altitude ;  hence 
the  sum  of  their  volumes,  that  is, 
the  volume  of  the  given  prism,  will 
be  equal  to  the  sum  of  their  bases 
multiplied  by  the  common  altitude 
(InTo.  108,  Cor.  Ill);  that  is,  the 
whole  volume  equals  the  whole  base 
multiplied  by  the  altitude.     Q.  E.  D. 


Through 


N 

; 

1 

/ 

\    / 

■^ 

/ 

\ 

■-•■/ 

\  / 

..  ••    / 

^ 

E 

k 

F 

- 

/■'. 

/  \ 

i 

..^ 

Y   '-•■ 

.••■■' 

/ 

^^ 

3 

c 

94 


ELEMENTS  OF 


PKOPOSITIOJST  IX. 

110.  Jf  a  jpyramid  he  cut  hy  a  jdane  parallel  to  its 
hase,  the  edges  and  the  altitude  will  he  divided  propor- 
tionally^ and  the  section  will  he  a  polygon  similar  to  the 


Proof.  Let  the  pyramid  Y—ABC^  wliose  altitude  is 
O  V,  be  cut  by  a  plane  ahc  par- 
allel to  the  base  ABC.  Then 
ah  and  AB,  he  and  BO,  ac  and 
A  Oy  as  also  ao  and  A  0,  will  be 
parallel  (IS'o.  90).  Hence  the 
angles  ahc  and  ABOy  hca  and 
BOA,  cah  and  CAB  are  respec- 
tively equal  (No.  93),  and  the 
two  polygons  are  similar.  The 
triangles  AOV  and  ao  V  are 
also  similar,  as  also  are  the  tri- 
angles ah  V  and  AB  V.     Hence 

ah:  AB::aV:  AV::ao  :  AO.     Q.  E.  D. 

Cor.  I.  The  area  of  any  section  parallel  to  the  hase  of 
the  pyramid  is  proportional  to  the  square  of  its  distance 
from  the  vertex.    For 


area  ahc  :  area  ABC  \\  aV  '.  AB"": 


ao 


AO' 


Cor.  II.  If  tvw  pyramids  stand  on  the  same  plane, 
and  have  equal  altitudes,  the  sections  made  through  them 
hy  a  'plane  parallel  to  that  of  their  hases  will  he  propor- 
tional to  the  hases.  For  if  B  and  B'  are  the  bases  of 
the  two  pyramids,  and  H  their  altitude,  and  if  h  and  h' 
are  the  two  sections,  and  h  their  distance  from  the  ver- 
tex, then 


GEOMETRY.  95 


B:h::ir  :  h\  and  B'  :h'::H'  :  h'; 

and  consequently       B  :  h  ::  B'  :  h' . 

Cor.  III.  If  two  jpyramids  have  the  same  altitude  and 
equivalent  bases,  any  sections  made  though  them  at 
equal  distances  from  the  hases,  or  from  the  vertices, 
will  have  an  equal  area. 


PEOPOSITION  X. 

111.  The  convex,  or  lateral,  surface  of  a  right  jpyra- 
mid  is  equal  to  half  the  product  of  the  slant  height  into 
the  perimeter  of  the  base. 

Proof.  In  a  right  pyramid  the  base  is  a  regular  poly- 
gon, and  the  lateral  faces  are  equal  isosceles  triangles. 
The  surface  of  one  of  these  triangles  is  equal  to  half  the 
product  of  its  base  into  its  altitude,  which  altitude  is  the 
slant  height  of  the  pyramid.  Hence  the  lateral  surface 
of  the  pyramid,  which  is  equal  to  the  sum  of  the  sur- 
faces of  these  triangles,  is  expressed  by  half  the  product 
of  the  perimeter  of  the  base  into  the  slant  height  of  the 
pyramid.     Q.  E.  J). 

Cor.  If  a  right  pyramid  be  truncated  by  a  plane  par- 
allel to  its  base,  the  convex  surface  of  the  remaining 
frustum  will  be  equal  to  the  sum  of  the  surfaces  of 
all  the  lateral  faces.  Such  faces  are  trapezoids,  and  each 
of  them  is  equal  to  half  the  product  of  its  altitude  into 
the  sum  of  the  parallel  sides.  Their  sum,  or  the  surface 
of  the  frustum,  is  therefore  half  the  product  of  tlie  slant 
height  into  the  sum  of  the  perimeters  of  the  parallel 


96 


ELEMENTS  OF 


PEOPOSITIOI^  XL 

112.  Two  triangular  pyramids  ha/ving  equal  alti- 
tudes and  equivalent  bases,  are  equal  hi  volume. 

Proof.  Let  Y-ABG  and  Y'-A'B'C  be  two  pyra- 
mids having  equal  altitudes  and  equivalent  bases.  Di- 
vide  their  altitude  into  equal  parts,  and  through  the 


points  of  division  pass  planes  parallel  to  the  plane  of 
the  bases.  The  sections  made  by  each  of  these  planes 
will  be  respectively  equal  (No.  110,  Cor.  III.) 

On  ABC,  DEF,  GHl,  KLM  as  lower  bases  con- 
struct as  many  exterior  prisms  between  the  parallel 
planes.  Their  altitudes  w411  be  equal.  On  D'E'F'^ 
G'H'I',  K'L'M'  as  upper  bases  construct  as  many  in- 
terior prisms.  They  will  have  equal  altitudes,  and  their 
bases  will  be  respectively  equal  to  the  bases  DEF^  GUI, 
KLM  of  the  prisms  in  the  other  pyramid.  They  will 
therefore  be  equal  each  to  each.  The  difference  between 
the  sum  of  the  exterior  prisms  made  in  the  pyramid 
Y^ABC  and  the  sum  of  the  interior  prisms  made  in 
the  pyramid  F'— ^'^'6^' is  therefore  the  prism  whose 
base  is  ABC.     This  difference  will  become  less  and  less 


GEOMETRY.  97 


in  proportion  as  our  prisms  have  a  less  and  less  altitude ; 
and  therefore  if  the  altitude  of  the  pyramids  be  divided 
into  an  indefinitely  increasing  number  of  parts,  and 
if  an  infinitely  increasing  number  of  exterior  and  inte- 
rior prisms  be  constructed,  having  indefinitely  decreas- 
ing altitudes,  the  difference  between  the  sum  of  the 
exterior  and  interior  prisms  will  be  a  prism  of  an  inde- 
finitely small  altitude  constructed  on  ABC  as  a  base ; 
that  is,  a  volume  less  than  any  assignable  volume.  But 
the  two  pyramids  evidently  differ  less  from  each  other 
than  the  two  sums  of  said  exterior  and  interior  prisms. 
Therefore  the  two  pyramids  have  volumes,  between 
which  no  difference  is  assignable ;  which  is  to  say,  that 
the  two  pyramids  are  equal  in  volume.     Q.  E.  D. 


PEOPOSITIOJS-   XII. 

113.  Any  triangular  jprism  can  he  divided  into  three 
equal  triangvla/r  pyramids. 

Proof.  Let  ABC—F  be  a  triangular  prism.  Draw 
the  planes  A  CE  and  DCE.  The 
prism  will  be  divided  into  three 
pyramids.  The  pyramids  J. ^(7 
—E  and  BEE—  C  are  evident 
ly  equal,  as  they  have  equal 
bases  and  altitudes.  The  pyra- 
mids ACB-E  and  BCE -E 
are  also  equal,  as  their  bases  are 
halves  of  the  same  parallelo- 
gram, and  their  vertex  is  at  the 
same  point  E.  Therefore  the  three  pyramids  are  equal 
in  volume.     Q.  E.  D. 

Cor.  I.  The  volume  of  a  triangular  jpyramiid  is  equal 
to  one-third  of  the  jproduct  of  its  hase  into  its  altitude. 


98 


ELEMENTS  OF 


Cor.  IT.  The  volume  of  any  pyramid  is  equal  to  one- 
third  of  the  product  of  its  hase  into  its  altitude.  For 
any  pyramid  can  be  divided  into  a  number  of  triangular 
pyramids,  whose  total  volume  will  be  one-third  of  the 
product  of  the  sum*  of  their  bases  into  their  common  al- 
titude ;  and  the  sum  of  their  bases  is  the  base  of  the 
pyramid  which  they  compose. 


PKOPOSITIOIsr  XIII. 

114.  The  volume  of  the  frustum  of  a  pyramid  is 
equal  to  the  sum  of  the  volumes  of  three  pyramids  whose 
common  altitude  is  the  altitude  of  the  frustum^  and 
whose  hases  are  the  lower  hase  of  the  frustum^  the  up- 
per hase  of  the  frustum,  and  a  mean  proportional  he- 
tween  them. 

Proof  'LQiABCDE-a  be 
the  frustum  of  a  pyramid 
whose  vertex  is  Y.  The  vol- 
ume of  this  frustum  is  equal 
to  the  volume  of  the  pyramid 
ABODE—  V  minus  the  vol- 
ume of  the  pyramid  ahcde  —  Y. 
Make  ABCDE=.P  and  ahcde 
=^,  also  YO—H,  and  Yo—h, 
and  call  F  the  volume  of  the 
frustum.  Then  (ISTo.  113,  Cor. 
11.) 

F=\{Py.E:-py.h). 

But  (No.  110,  Cor.  I.) 


!P 


P'.:h':  H';  hence  j?=P 


Z- 
H' 


GEOMETRY.  99 


and  therefore 


But  H'-h'^  {S-h)  (jy'+  ^A  +  h?) ; 

and  consequently 

■  H-h  (P.H'+  PHh + PA'\ 


or 


F= 


7a  7 

I^ow,  P-^^=:^,  and  P-^  is  a  mean  proportional  be- 

tween  P  and  P-^a  ?  that  is,  between  P  and  ^.     There- 
fore 


and  since  H—h  is  the  altitude  of  the  frustum,  it  follows 
that  the  volume  of  the  frustum  is  equal  to  the  sum  of 
the  volumes  of  three  pyramids  of  the  same  height,  whose 
bases  are  the  lower  and  the  upper  bases  of  the  frustum, 
and  a  mean  proportional  between  them.     Q.  E.  D. 


PKOPOSITIO]^  XIY. 

115.  Similar  triangular  prisms  are  to  one  a/nother 
as  the  cubes  of  their  homologous  edges  or  linear  dimen- 
sions. 

Proof.  The  prisms  ABC—Fdm^  ahc—f  if  similar, 


100 


ELEMENTS  OF 


have  similar  faces  between  homologous  edges,  equally 


inclined  to  their 
Hence,  letting  fall  DO  and 
do  perpendicular  to  ^^6^ 
and  dbc,  and  drawing  AG 
and  ao,  the  triangles  A  OD 
and  aod  will  be  similar; 
for,  besides  the  right  an- 
gles at  0  and  <?,  thev  have 
the  angles  Z>J.6^  and  dao  equal, 


Consequently 


DO\do\\AD\ad. 

On  the  other  hand,  from  the  similarity  of  the  figures  we 
have 

ABC\  ahcwAR  :  al^wAD';  '^ ; 

hence,  multiplying  these  proportions  term  by  term,  we 
have 

ABCxJyO  :  abcxdo  ::  AD' :  S'/ 

but  ABCxDO  and  aboXdo  are  the  volumes  of  the  two 
prisms;  therefore 

Prism  ABC-F :  Prism  abc  -/ : :  AD'  :  ad".    Q.  E.  D. 

Cor,  I.  Any  two  similar  jprisms  are  to  each  other  as 
the  cubes  of  their  homologous  dimensions.  For  their 
bases  are  similar  polygons  containing  the  same  number 
of  similar  triangles,  similarly  placed ;  and  consequently 
these  prisms  consist  of  similar  triangular  prisms,  which 
are  to  each  other  as  the  cubes  of  their  homologous  di- 
mensions ;  and  as,  for  equal  ratios,  the  sum  of  the  ante- 
cedents is  to  the  sum  of  the  consequents  as  each  antece- 
dent is  to  its  consequent,  the  whole  volume  of  one  prism 


GEOMETRY. 


101 


is  to  the  whole  volume  of  the  other  as  each  triangular 
prism  is  to  its  similar  one. 

Cor.  II.  Similar  jpyramids  are  to  each  other  as  the 
cubes  of  their  homologous  dimensions.  For  they  are 
similar  parts  of  similar  prisms. 


PROPOSlTIOISr  XY. 

116,  The  volume  of  a  wedge  is  equal  to  the  volume 
of  a  pyramid  having  for  its  hase  a  right  section  of  the 
wedge,  perpendicular  to  its  hach,  and  an  altitude  equal 
to  the  sum  of  its  three  parallel  dimensions. 

Proof.  The  wedge  is  a  volume  bounded  by  a  rectangle 
ABCD  called 

the   hacJc,   two  E F 

equal  trape- 
zoids ABEF, 
DGEF  called 
faces,  and  two 
equal  triangles 
AED,  BEG 
called  ends. 
The  line  EF, 
in    which    the 

faces  meet,  is  called  the  edge.  The  volume  is  composed 
of  three  parts,  viz.,  of  a  rectangular  prism  comprised  be- 
tween the  two  sections  EGH,  FKL,  perpendicular  to 
the  back  of  the  wedge,  and  of  the  two  equal  pyramids 
AHGD-E  and  LBGK-F. 

Let  AB—l,  EE^l\  AD=h,  and  the  altitude  EI  of 
the  triangle  EGH  (which  is  also  the  altitude  of  the 
wedge)  =  h.  The  expression  of  the  volume  Y  of  the 
wedge  will  be 


102 


ELEMENTS  OF 


V=  prism  GIIE-F-\-  two  pyramids  AHGB  -E. 

Now,  the  volume  of  the  prism  is  \bhy,HL  —  \hlil' ;  and 
the  volume  of  the  two  pyramids  is  \hy^AD  {AII-\-LB) 
=  \hh  {l-V),  Therefore  F=  ^l)hV-{-  \bh  {l-V) ;  which 
may  be  put  under  the  form 


2  V     3    7 ' 


and  this  is  the  volume  of  the  pyramid  above  mentioned. 
Q.  E.  D. 

When  the  length  I  of 
the  back  is  less  than  the 
length  V  of  the  edge,  the 
volume  will  be  equal  to 
that  of  the  right  prism 
EGII-Fmhms  that  of 
the  two  equal  pyramids 
JIABG-Esind  BLKC 
—F.    And  therefore 

which  again  may  be  reduced  to  the  form 

And  the  same  result  applies  also  to  a  wedge  in  which 
I  =  V .     For  then  we  have 


y-'i^^)-^^^^ 


which  is  the  expression  of  the  volume  of  the  rectangular 
prism  whose  base  is  — ,  and  whose  altitude  is  I. 


GEOMETRY, 


103 


IK 


PEOPOSITIOlSr  XYI. 

117.  The  volume  of  a  prismoid  is  equal  to  the  vol- 
ume of  a  pyramid  of  the  sam^e  altitude,  having  for  hase 
half  the  sum  of  its  parallel  faces,  plus  twice  the  middle 
section  hetween  the  parallel  faces. 

Proof  A  prismoid  is  a  frustum  of  a  wedge.  Its  faces 
ABCB  and  EFGH  2.yq  paral- 
lel rectangles ;  and  its  altitude 
is  KI  perpendicular  to  the 
parallel  faces.  Drawing  a  dia- 
gonal plane  BGFE,  the  vol- 
ume of  the  prismoid  will  be 
divided  into  two  wedges ;  and 
therefore,  putting  AB  =  I, 
EF=.l\  DA  =  h,  EII=l', 
and  Kl  =  A,  we  shall  have  (ISTo.  116) 

^    hh/2i+r\  .  h'h/2i'+i\ 


or 


Y=^  (2U  +  2bT+  U+  1% 


which  may  be  written  thus : 

F=  ^  {U  -f  l'l'+  M-{-  VT)  +  ^  (J^  +  hT). 

Draw  the  middle  section  LMON,  and  make  LM  =  m^ 
LN  =  n.  As  this  section  is  midway  between  the  bases, 
we  have- 

2m  =  Z-f  Z',  271  =  l  +  h\  and  ^mn  =  {l^l')  {h-\-l)% 
or  -  4m7i  =  Z5  +  VV+  W+  Vh  ; 


104  ELEMENTS  OF 


Substituting  this  in  the  expression  of  the  volume,  we 
have 


or 


which  is  the  volume  of  the  pyramid  above  mentioned. 
Q.  E.  D. 


GEOMETRY,  105 


BOOK   YII 


8UBFACE8   AND    VOLUMES   OF   ROUND 
BODIES. 

118.  Round  bodies  are  those  geometric  bodies  wbicb 
can  be  described  by  a  plane  figure  revolving  about  a 
fixed  line  called  the  axis  of  revolution.  Thus  a  right- 
angled  triangle  revolving  about  one  of  its  sides  describes 
a  cone  /  a  rectangle  revolving  about  one  of  its  sides  de- 
scribes a  cylinder ;  a  semicircle  revolving  about  its  dia- 
meter describes  a  sphere,  and  so  forth.  As  every  point 
of  the  revolving  figure  describes  the  circumference  of  a 
circle,  the  determination  of  the  surfaces  and  volumes  of 
round  bodies  must  be  based  on  the  relations  existing  be- 
tween straight  lines  and  circular  curves.  The  unit  of 
measure  for  all  surfaces  is,  of  course,  a  square  construct- 
ed on  a  unit  of  length,  and  the  unit  of  measure  for  all 
volumes  is  a  cube  constructed  on  the  unit  of  length ;  this 
unit  of  length  being  always  a  straight  line.  Now,  round 
surfaces  and  volumes  cannot  be  measured  by  any  straight 
unit,  unless  they  be  considered  as  made  up  of  a  multi- 
tude of  straight  consecutive  elements,  a  little  inclined  to 
one  another.  Hence,  in  treating  of  round  bodies,  we 
must  keep  in  mind  that  it  is  only  by  regarding  the  circle 
as  a  regular  polygon  having  an  exceedingly  great  multi- 
tude of  sides  exceedingly  small,  that  its  area  and  its  cir- 
cumference become  expressible,  in  terms  of  rectilinear 
measures,  to  any  desired  degree  of  approximation,  as  we 
have  explained  above  (IsTos.  60,  61). 


106  ELEMENTS  OF 


PEOPOSITION  I. 

119,  The  convex  surface  of  a  right  cylinder  is  equal 
to  the  product  of  the  circumference  of  its  hase  into  its 
altitude. 

Proof.  A  right  cylinder  is  what  a  right  prism  be- 
comes when  its  base  is  a  regular  poly- 
gon having  an  innumerable  multitude 
of  sides.  Its  convex  surface  is  there- 
fore expressed,  like  that  of  every  right 
prism  (No.  102),  by  the  product  of  its 
altitude  into  the  perimeter  of  its  base, 
the  perimeter  being,  in  this  case,  the  cir- 
cumference of  a  circle.     Q.  E.  D. 

Let  AB=^h  be  the  altitude  of  the  cylinder,  and  CA 
=  r  the  radius  of  the  base.  Then  the  convex  surface 
willbe/(S=2;r7'X>^. 


PKOPOSITIO]?^  II. 

120.  The  volume  of  a  right  cylinder  is  equal  to  the 
jproduct  of  the  area  of  the  hase  into  the  altitude. 

Proof  The  volume  of  any  right  prism  is  equal  to  the 
product  of  its  altitude  into  the  area  of  its  base  (No.  109). 
But  a  right  cylinder  is  nothing  but  a  right  prism  whose 
regular  base  has  an  innumerable  multitude  of  sides. 
Therefore  the  volume  of  a  right  cylinder  is  equal  to 
the  product  of  its  altitude  into  the  area  of  its  base. 
Q.  E.  D. 

Let  h  be  the  altitude,  and  r  the  radius  of  the  base ; 
then  the  volume  of  the  cylinder  is  Y—  itr'  X  A. 


GEOMETRY.  107 


PKOPOSITION  III. 

121.  The  convex  surface  of  a  right  cone  is  equal  to 
half  the  product  of  the  circumference  of  its  hase  into  its 
slant  height. 

Proof  The  convex  surface  of  any  right  pyramid  is 
equal  to  half  the  product  of  the  peri- 
meter of  its  base  into  its  slant  height 
(N"o.  111).  But  the  right  cone  is  no- 
thing but  a  regular  right  pyramid 
whose  base  has  a  circumference  for  its 
perimeter.  Therefore  the  convex  sur- 
face of  a  cone  is  equal  to  half  the  pro- 
duct of  the  circumference  of  its  base 
into  its  slant  height.     Q.  E.  D. 

If  AB  =  7'  is  the  radius  of  the  base,  and  B  V=  s  the 
slant  height,  the   convex   surface  of  the  cone  will  be 

S=2Kr  X  ^=^7:rXs. 


PKOPOSITIOX  lY. 

122.  The  volume  of  a  cone  is  equal  to  the  product 
of  the  area  of  its  hase  into  one-third  of  its  altitude. 

Proof  The  volume  of  a  right  pyramid  is  equal  to  the 
product  of  the  area  of  its  base  into  one-third  of  its  alti- 
tude (No.  113)  ;  and  the  cone  is  a  right  pyramid  whose 
base  is  a  regular  polygon  of  an  innumerable  multitude 
of  sides.     Hence  the  conclusion.     Q.  E.  D. 

Let  h  be  the  altitude  of  the  cone,  and  r  the  radius  of 
its  base ;  then  its  volume  will  be  F  =  ^nr"^  X  h. 


108  ELEMENTS  OF 


PEOPOSITIO]Sr  Y. 

123.  The  convex  surface  of  the  frustum  of  a  cone 
is  equal  to  half  the  product  of  its  slant  height  into  the 
sum  of  the  circumferences  of  the  parallel  hases. 

Proof  The  convex  surface  of 
the  frustum  of  a  pyramid  is  e^jual 
to  half  the  product  of  its  slant 
height  into  the  sum  of  the  peri- 
meters of  the  parallel  bases  (No. 
Ill,  Cor.)  When  the  perime- 
ters become  circumferences  of 
circles,  the  frustum  of  the  pyra- 
mid becomes  the  frustum  of  a 
cone.     Hence,  etc.     Q.  E.  D. 

Let  r  and  r'  be  the  radii  of  the  bases,  and  s  the  slant 
height ;  then  the  convex  surface  of  the  frustum  will  be 
S^r.(T-\-T')s. 


PEOPOSITIOlSr  YI. 

124.  The  volume  of  the  frustum  of  a  cone  is  equal 
to  the  sum  of  the  volumes  of  three  cones  whose  common 
altitude  is  the  altitude  of  the  frustum,  and  whose  hases 
are  the  lower  hose  of  the  frustum,  the  upper  hase  of  the 
frustum,  and  a  mean  proportional  between  them. 

Proof  Tliis  proposition  has  been  proved  (No.  114) 
for  any  frustum  of  a  pyramid.  And  therefore  the  same 
is  true  for  the  frustum  of  a  cone,  which  is  nothing  but 
the  frustum  of  a  right  pyramid  whose  base  is  a  regular 
polygon  of  an  infinite  multitude  of  sides.     Q.  E.  D. 


GEOMETRY. 


109 


Let  T  and  r'  be  the  radii  of  the  two  bases,  and  h  the 
altitude  of  the  frustum.     Its  volume  will  be 


PEOPOSITIOJST  YII. 

125.  If  a  regular  semi-polygon  he  revolved  about  its 
axis,  the  surface  generated  hy  the  semi-^jperimeter  will  he 
equal  to  thejproduct  of  the  axis  into  the  circumference 
of  the  inscribed  circle. 

Proof.  Let  ABGDEF  be  a  regular  semi-polygon, 
AF\i'&  axis,  and  OM  its  apothem,  or  the  radius  of  the 
inscribed  circle.  From  the  extremi- 
ties of  any  side,  as  BC,  draw  BG 
and  CH  perpendicular  to  AF,  and 
BN  perpendicular  to  CH.  The  sur- 
face generated  by  ^6^  is  the  convex 
surface  of  the  frustum  of  a  cone, 
whose  bases  are  27rX^-^and  27rX 
BG,  and  whose  slant  height  is  BG ; 
hence  such  a  surface  (Ko.  123)  is 
equalto^(7X7r((7^+^6^).  From 
the  middle  point  Jf  of  BG  draw 
Ml  perpendicular  to  AF ;  then 
GH^BG  —  '^MI,^^^  the  expres- 
sion of  the  surface  becomes 

BGy.  2nMI. 

Now,  from  the  similar  triangles  GBJSF  and  MGl  we 
have 

BG :  BJST: :  GM :  MI: :  2;r6)Jf :  2;rJ[//; 


110  ELEMENTS  OF 


hence  BCy.'I'kMI^BNy^'ItiOM ;  - 

but  BNz=.  GH ;  therefore  the  generated  surface  will  be 
GEy.'lnOM; 

that  is,  the  surface  generated  by  any  side  is  equal  to  the 
product  of  the  projection  of  that  side  on  the  axis  into 
the  circumference  of  the  inscribed  circle.  Hence  the 
surface  generated  by  the  entire  semi-perimeter  is  equal 
to  the  product  of  the  sum  AF  of  the  projections  of  its 
sides  into  the  circumference  of  the  inscribed  circle. 
Q.  E.  D. 

Cor.  The  surface  generated  hy  any  portion  of  the 
senni-jperimeter  is  equal  to  the  jproduct  of  its  projection 
into  the  drcumference  of  the  inscribed  circle. 


PEOPOSITION  YIIL 

126.  The  surface  of  a  sphere  is  equal  to  the  product 
of  its  diameter  into  the  circumference  of  a  great  circle. 

Proof.  The  surface  of  the  sphere  is  generated  by  the 
revolution  of  a  semi- circumference  around  its  diameter. 
Now,  a  semi-circumference  is  nothing  but  a  regular  semi- 
perimeter  of  a  numberless  multitude  of  sides ;  its  projec- 
tion is  its  diameter;  and  the  radius  of  the  inscribed 
circle  is  nothing  else  than  its  own  radius.  Therefore 
(No.  125)  the  surface  of  the  sphere  is  equal  to  the  pro- 
duct of  its  diameter  into  the  circumference  of  a  circle  of 
the  same  diameter.     Q.  E.  D. 

Cor.  1.  The  surface  of  the  sphere  whose  radius  is  r  is 
Sz=^^Tzry,%r-=.^:Tzr^.  It  is  therefore  equal  to  four  times 
the  area  of  a  great  circle. 


GEOMETRY. 


Ill 


Cor.  II.  The  surface  gene- 
rated by  any  arc  of  a  semi- 
circle, is  equal  to  the  product 
of  its  projection  on  the  diam- 
eter into  the  circumference  of 
a  circle  of  the  same  diameter 
(No.  125,  Cor.)  Such  a  sur- 
face is  called  a  zone.  Thus,  if 
the  arc  AG  revolves  around 
the  diameter  MN^  it  generates 
the  zone  A  BCD,  whose  alti- 
tude is  EF,  and  whose  area  is  ^izry^EF. 

Cor.  III.  Two  zones  on  the  same  sphere  are  to  each 
other  as  their  altitudes. 


PEOPOSITION  IX. 

1!37.  The  volume  generated  by  a  triangle  revolving 
around  its  hase  is  equal  to  the  jproduct  of  one-third  of 
its  base  into  the  area  of  the  circle  described  by  the  vertex 
opposite  the  base. 

Proof.  Let  the  triangle  ABC vqyoIyq  around  its  base 
^^  as  an  axis.  Draw  from  C 
the  line  CD  perpendicular  to  the 
axis.  The  portion  ADC  of  the 
triangle  will  generate  a  cone 
whose  volume  (No.  122)  will  be 

\n  .  CD'XAD, 

and  the  portion  BCD  will  generate  a  cone  whose  vol- 
ume will  be 

\n.CD'xDB. 


112 


ELEMENTS  OF 


Hence  the  total  volume  generated  will  be 

\7i,  CD'  {AD-\-DB\  or  \7iCD'X  AB, 

If  the  perpendicular  CD  falls  on  the  prolongation  of 
the  base  AB,  the  result  is  still  the  same.  For  in  this 
case  the  volume  generated  bj  the  triangle  ABC  will  be 
the  difference  between  the  volumes  of  the  cones  gene- 
rated bj  A  CD  and  BCD  ;  that  is, 

\7: .  CD'  {AD-DB\  or  \'kCD' X  AB,    Q.  E.  D. 


PKOPOSITIOlSr  X. 

128.  If  an  isosceles  triangle  revolves  about  a  line 
passing  through  its  vertex,  the  volume  generated  will  he 
equal  to  the  jproduct  of  one-third  of  the  altitude  of  the 
triangle  into  the  surface  generated  by  its  base. 

Proof.  Let^^  be  the  base,  (7  the  vertex,  and  6Tthe 
altitude  of  an  isosceles 

triangle,   and    suppose,  ' 

first,  the  base  AB  to  be 
parallel  to  the  axis  of 
revolution-  CD.  Draw 
AP  and  BQ  perpendi- 
cular  to  the  axis.     The 

volume  generated  by  the  triangle  will  be  equal  to  the 
volume  of  the  cylinder  generated  by  the  rectangle 
APQB,  minus  the  volumes  of  the  two  cones  generated 
by  the  two  equal  triangles  APC  and  BQC.  The  vol- 
ume of  the  cylinder  is  tiCPx  AB,  and  that  of  the  two 

cones  is  ^nAl^'X  PC+irrBQ'X  CQ,  or  IttCI'X  AB. 
Hence  the  volume  generated  by  the  triangle  ABC  is 


GEOMETRY. 


113 


F=|;r(7Px  AB=\CIxAB  X  ^nCI, 

where  AB  X  2yr(7/  expresses  the  surface  generated  by 
the  base  AB, 

Suppose,  in  the  second 
place,  the  base  AB  to  be 
inclined  to  the  axis  CD. 
Produce  AB  till  it  meets 
the  axis  at  D.     The  vol- 
ume  generated    by   the 
triangle  ABC  will   be 
equal    to    the     volume     q 
generated  by  the  trian- 
gle A  CD  minus  the  volume  generated  by  the  triangle 
BCD.     We  have  therefore   (Ko.  127),  AE  and  BF 
being  the  altitudes  of  these  two  triangles, 

Y=^7:AE'  XCD-  \'kBF'  X  CD, 

that  is,       Y=.\tz  {AE'-  BE')  X  CD. 

Draw  IG  perpendicular  and  BH  parallel  to    CD. 
Since 

AE'-  BF'^  {AE^  BF)  {AE-  BF)  =  2IG  x  AR, 

the  expression  of  the  volume  becomes 

V=i7:.JG.AR.CD. 

But  the  right-angled  triangles  ABII  and   CID,  which 
are  similar,  give  AH :  AB  ::  CI :  CD  •  hence 

AH.CD  =  AB  .CI; 

and  consequently 

Y=\Cly.'^7zlGxAB, 
where  ^nlGxAB  is  the  surface  generated  by  AB, 


114  ELEMENTS  OF 


Suppose,  tliirdlj,  tlie  base  f^ 

AB  to  be  so  inclined  that  /\ 

the  point   B  falls  on  the  /      \\ 

axis   CD.     Draw  AE  and  /^--•'^T  iX 

IG  both   perpendicular   to     ^^^ ^  ^  |^ g- 

CD.     The  volume  generat- 
ed by  the  revolution  will  be  (No.  12Y) 

Y=:^jcAE'X  CB. 

But,  since  the  triangles  AEB  and  CIB  are  similar,  we 
have 

AE  \AB\\C1\  CB;  hence  AE .  CB  ^  AB  .  CI. 

Therefore  F=  \7zAE .  AB  .  CI. 

But  AE=  21  G ;  and  consequently  * 

V=\Clx27:IGxAB, 

where  ^nlG  X  AB  is  the  surface  generated  by  AB, 
Hence  the  theorem  enunciated  is  true  in  all  cases. 
Q.E.D. 


PEOPOSITIOISr  XL 

129.  The  volimie  generated  hy  a  regular  semi-poly- 
gon revolving  about  its  axis  is  equal  to  the  product  of 
one-third  of  its  apothem  into  the  surface  generated  hy 
the  semi-perimeter. 

Proof  Let  ABCDEEhe  a  regular  semi-polygon,  and 
01  its  apothem.  If  we  draw  BO,  CO,  ...  so  as  to  di- 
vide it  into  isosceles  triangles,  each  triangle  will,  by  its 


GEOMETRY. 


115 


revolution,  generate  a  volume  equal 
to  the  product  of  one-third  of  its  al- 
titude into  the  surface  generated  by 
its  base  (No.  128).  And,  since  all 
these  triangles  have  for  their  altitude 
the  apothera  of  the  semi-polygon,  the 
total  volume  generated  will  be  ex- 
pressed by  the  product  of  one-third 
of  the  apothem  into  the  whole  sur- 
face generated  by  the  semi-perimeter. 
Q.  E.  D. 

Cor.  The  volume  generated  by  any 
portion  OB  CD  of  the  semi-polygon 
is  equal  to  the  product  of  one-third 
of  the  apothem  into  the  surface  generated  by  the  corre- 
sponding portion  of  the  semi-perimeter. 


PEOPOSITION   XII. 

,130.  TJie  volume  of  a  sphere  is  equal  to  the  j^roduct 
of  its  surface  into  one-third  of  its  radius. 

Proof.  The  volume  of  the  sphere  is  generated  by  the 
revolution  of  a  semicircle  about  its  diameter.  Kow,  a 
semicircle  is  nothing  but  a  regular  semi-polygon  of  a 
numberless  multitude  of  sides,  whose  apothem  is  nothing 
else  than  its  own  radius.  Therefore  (No.  129)  the  vol- 
ume of  the  sphere  is  equal  to  the  product  of  one-third  of 
its  radius  into  the  whole  surface  of  the  sphere.     Q.  E.  D. 

Cor.  1.  The  surface  of  the  sphere  being  ^nr^  (No.  126, 
Cor.  I.),  its  volume  will  be 


F: 


~3~' 


116 


ELEMENTS  OF 


Cor.  11.  Any  circular  sector,  that  is,  any  portion  of  a 
semicircle  bounded  by 
two  radii,  will  generate 
a  volume  equal  to  the 
product  of  one-third  of 
the  radius  into  the  sur- 
face, or  zone,  generated 
by  the  arc  of  the  sector. 
Such  a  voliJme  is  a 
spherical  sector.  Thus 
ABODG  is  a  spherical 
sector  generated  by  the 
revolution  of  the  circular 
sector  GAB  about  the 
diameter  EF.  The  por- 
tion AOG oi  the  sphere  forms  no  part  of  the  volume  of 
this  spherical  sector. 

Cor.  III.  The  volumes  of  different  spheres  are  to  one 
another  as  the  cubes  of  their  radii,  or  of  their  diameters. 


PEOPOSITIOE"  XIII. 

131.  The  volume  generated  hy  a  circular  segment 
revolving  about  a  diameter  exterior  to  it,  is  equal  to  a 
cone  having  for  its  hase  tlie  circle  constructed  on  the 
chord  of  the  segment  as  a  diameter,  and  for  its  altitude 
tivice  the  j^ojection  of  that  chord  on  the  axis  of  revo- 
lution. 

Proof.  Let  A  CB  be  the  segment  of  a  circle.  Bisect 
the  chord  AB  in  D  ;  draw  the  radii  AG  and  BG,  and 
let  fall  the  perpendiculars  AH,  BG,  and  DI  on  the  axis 


GEOMETRY. 


iir 


EF ;  draw  also  ^iV^  perpendicular  to  BG,  and  OD  per- 
pendicular to  the  chord  AB.  The 
volume  generated  by  the  segment 
A  CB  will  evidently  be  equal  to  the 
volume  generated  by  the  sector 
ACBO,  minus  the  volume  generat- 
ed by  the  isosceles  triangle  ABO. 
Hence  (ISTos.  130  and  128) 

Y=27:AOxGHx^AO 

-^TiDIxABx^BO. 

But  from  the  similar  triangles  OBI 
and  BAN  we  have 

Dl\  OD::AN:AB, 
whence 

JDIx  AB=ODx  A]Sr=  OB  x  GH ; 

hence,  by  substitution  and  reduction, 

F=  |;r .  GR  {AG'-  BO')  =  \7z.GH,  AB% 
^      2GH 


or 


X  7rAB\ 


which  is  the  volume  of  a  cone  whose  altitude  is  ^GH 
and  whose  base  has  the  radius  AB.    Q.  E.  D. 


PKOPOSITION^  XIY. 

132.  The  volume  of  a  sjpherical  segment  is  equal  to 
the  volume  of  a  sphere  whose  diameter  is  the  altitude  of 
the  segment^  plus  the  volumes  of  two  cylinders  whose 
bases  are  the  bases  of  the  segment,  and  whose  altitude 
is  half  that  of  the  segment. 

Proof.  Let  A  CB  be  the  arc  of  a  circle.     Bisect  the 


118 


ELEMENTS  OF 


chord  AB  in  D.     Draw  AH,  BG, 

and  7>/ perpendicular  to  the  diame- 
ter SF.  If  the  area  ACBGH  re- 
volves  about  JEF.  it  will  generate  a 
spherical  segment,  whose  volume  will 
consist  of  the  volume  generated  by 
the  circular  segment  ACB,  and  of 
the  conical  frustum  generated  by  the 
trapezoid  ABGH.  Hence  (Kos. 
131  and  124) 

2GH 


V: 


TT  .AD' 


GH 


+  ^Tzi^AW^BG'-^^AIIy^BG), 

!N'ow,  since  AB^'^AB,  we  have 

AB'=:  iAB'=  i  {AW'+  BIT'), 
and  because 

Jjr'=  GH'  and  BW'=  BG'+AH'-  2BG  X  AH, 

hence  AB'=i  {GH'+AH'+BG'-  ^AHxBG). 

Substituting  this  value  of  AB'  in  the  expression  of  V, 
we  find 

^      GH    /GH'+AH'+BG' 

^AHxBG+AH'-l-BG'-AHxBG^ ; 

which,  by  reduction,  becomes 

-^      GH    /SA7r+SB~G'+GIB\ 


GEOMETRY.  119 


whence,  putting  26^^/  instead  of  GH,  we  shall  finally 
obtain 

AttGI*  

y^  —3—  +  TzAH'x  GI+7cBG'x  GI, 

which  is  the  sum  of  the  volumes  designated  in  our  pro- 
position.    Q.  E.  D. 

Cor.  If  the  spherical  segment  has  but  one  base,  then 
AH  becomes  zero,  and  Gil  becomes  GE.  Hence  the 
expression  of  the  volume  will  become 

that  is,  a  spherical  segment  having  hut  one  hase  is  equal 
to  a  sphere  whose  diameter  is  the  altitude  of  the  segment, 
plus  a  cylinder  having  for  its  hase  the  hase  of  the  seg- 
ment, and  for  its  altitude  half  the  altitude  of  the  seg- 
m£nt. 


120  ELEMENTS  OF 


BOOK   YIII 


8PHEBICAL    GEOMETRY, 

133.  In  spherical  geometry  we  have  to  deal  with  an- 
gles, triangles,  and  polygons  the  sides  of  which  are  arcs 
of  circles  whose  planes  pass  through  the  centre  of  the 
sphere,  and  which  are  styled  great  circles.  The  angle 
made  by  two  arcs  of  great  circles  at  their  intersection  is 
a  dihedral  angle ;  for  it  measures  the  inclination  of  the 
planes  of  the  two  circles.  A  spherical  polygon  is  a  por- 
tion of  the  surface  of  the  sphere  bounded  by  arcs  of  a 
great  circle.  If  bounded  by  three  arcs  only,  it  will  be  a 
spherical  triangle  ;  if  bounded  by  two  arcs  only,  it  will 
be  a  lune^  and  the  two  arcs  will  then  be  two  semi-circum- 
ferences. The  solid  bounded  by  a  lune  and  two  semi- 
circles meeting  in  a  common  diameter  is  called  a  spheri- 
cal wedge.  A  spherical  pyramid  is  a  solid  bounded  by  a 
spherical  polygon  and  sectors  of  great  circles,  the  centre 
of  the  sphere  being  the  'vertex,  and  the  polygon  the  hase 
of  the  pyramid.  Such  are  the  main  subjects  to  be  inves- 
tigated in  the  present  book. 


PEOPOSITION  I. 

134.  The  angle  formed  hy  two  arcs  of  great  circles 
is  equal  to  that  formed  hy  their  tangents  at  their  jpoint 
of  intersection. 

Proof.  Let  the  angle  MAN  be  formed  by  the  two 


GEOMETRY. 


121 


arcs  AM  and  AI^.  The  tangent  AP  drawn  in  the 
plane  A  OJV,  and  the  tangent  A  Q 
drawn  in  the  plane  A  OM^  are  both 
perpendicular  to  the  radius  OA 
which  lies  in  the  intersection  of  the 
two  planes.  Hence  the  angle  of 
the  two  tangents  is  the  measure 
(No.  81)  of  the  inclination  of  the 
two  planes.  It  is  therefore  the 
measure  of  the  angle  MAN  deter- 
mined by  the  arcs  lying  in  those 
planes.     Q.  E.  D. 

Cor.  I.  The  arc  MW  of  a  great 
circle  perpendicular  to  ^6^  is  also 
the  measure  of  the  angle  MAN. 
For,  draw  MO  and  WO  ;  these  lines  will  be  parallel  to 
AQ  and  AP  respectively.     Hence  the  angle  MON  is 
equal  to  the  angle  QAP ;  and  therefore  the  arc  MN  is 
their  common  measure. 

Cor.  II.  The  vertical  angles  formed  at  the  intersection 
of  two  arcs  are  equal,  for  they  are  measured  by  the  ver- 
tical angles  of  their  tangents,  which  are  equal. 


PEOPOSITIOK  II. 

135.  Any  section  of  a  sphere  made  hy  a  plane  is  a 
circle. 

Proof.  Let  EFD  be  a  plane  section  of  the  sphere. 
Draw  the  radius  OP  perpendicular  to  the  plane  of  the 
section,  and  let  C  be   the  point  where  it  pierces   the 


122 


ELEMENTS  OF 


plane.  From  the  centre  of  the  sphere  draw  radii  to  any 
points  of  the  section, 
as  D  and  F.  The  tri- 
angles 6>(7i>  and  6>  67^ 
will  have  CO  common, 
OB  and  Oi^ equal,  and 
the  angle  in  C  right. 
Hence  (No.  27,  Cor.) 
the  third  side  CD  of 
the  one  will  be  equal 
to  the  third  side  CF 
of  the  other.  The 
point  C  is  therefore 
equally  distant  from 
any   points    taken    in 

the  section ;    and  consequently  the  section  is  a   circle. 
Q.  E.  D. 

Cor.  I.  The  section  is  greater  or  smaller  according  as 
it  is  nearer  to,  or  farther  from,  the  centre  of  the  sphere. 
The  greatest  section  passes  through  the  centre  of  the 
sphere  and  is  called  a  great  circle;  all  the  other  sec- 
tions are  called  sinall  circles. 

Cor.  II.  Every  great  circle  divides  the  sphere  and  its 
surface  into  two  equal  parts. 

Cor.  111.  The  centre  of  the  sphere  and  the  centre  of 
any  small  circle  are  in  a  straight  line  perpendicular  to 
the  plane  of  the  circle.  Thus  the  diameter  PQ  which 
passes  through  the  centre  of  any  circle  EFD  is  perpen- 
dicular to  its  plane,  and  the  points/^  and  Q  are  ih.^ j^oles 
of  the  circle ;  for  a  pole  of  a  circle  is  a  point  on  tlie  sur- 
face of  the  sphere  equally  distant  from  all  the  points  of 
the  circumference  of  the  circle.  Now,  this  is  the  case 
with  the  points  P  and  Q;  for,  if  we  conceive  chords  to 
be  drawn  from  P  to  i>,  to  E^  to  F^  etc.,  all  these  chords 


GEOMETRY.  123 

will  be  equal,  as  being  hypotenuses  of  equal  triangles, 
and  consequently  all  the  arcs  PB,  PE^  PF^  .  .  .  will 
also  be  equal.  And  in  a  similar  manner  the  arcs  QP,  QE^ 
QF^  .  .  .  will  be  equal. 

Cor.  1 V.  The  poles  of  a  great  circle  lie  at  90  degrees 
from  every  point  of  the  great  circle;  and  any  arc  PGQ 
passing  through  the  poles  of  a  great  circle  AGB  is  per- 
pendicular to  the  great  circle  (E^o.  95);  and  similarly 
any  arc  perpendicular  to  a  great  circle  passes  through  its 
poles. 

PKOPOSITIOI^  III. 

136.  Any  side  of  a  spherical  triangle  is  less  than  the 
sum  of  the  other  two. 

Proof  Let  ^^6^  be  a  spherical  triangle  situated  on  a 
sphere  whose  centre  is  0. 
Draw  OA,  OB,  00.  The 
trihedral  angle  0  will  be 
composed  of  three  plane 
angles,  of  which  any  one 
will  be  less  than  the  sum 
of  the  other  two  (^o.  98). 

And  since  these  angles  are  measured  by  the  arcs  AB, 
AC,  BC,  any  of  these  arcs  is  less  than  the  sum  of  the 
other  two.     Q.  E.  D. 


PROPOSITION  lY. 

13  7*  If  from  the  vertices  of  a  spherical  triangle,  as 
poles,  arcs  of  great  circles  he  descrihed  forming  a  spheri- 
cal triangle,  the  vertices  of  this  second  triangle  will  he 
poles  of  the  sides  of  the  first  triangle. 

■   Proof.  Let  ^^6^  be  the  given  triangle.    From^,^, 


124 


ELEMENTS   OF 


C  as  poles  describe  the  arcs  EF,  FD^  DE.  Since  A  is 
the  pole  of  EF^  the  dis- 
tance AE  is  a  quadrant; 
and  since  C  is  the  pole  of 
DE^  the  distance  CE  is  a 
quadrant  also ;  hence  the 
point  E  is  at  the  distance 
of  a  quadrant  from  the  arc 
A  C,  and  therefore  it  is  the 
pole  of  this  arc  (No.  135, 
Cor.  ly.)  It  may  be  proved 
in  the  same  manner  that  F 
is  the  pole  of  AB,  and  D  the  pole  of  BG.     Q.  E.  D 


PEOPOSITIOJSr  Y. 

138.  Any  angle  of  a  spherieal  triangle  is  measured 
hy  the  sujpjplement  of  the  side  ojyjposite  to  it  in  its  jpolar 
triangle. 

Proof.  Let  ABCh%  the  given  triangle,  and  DEF  its 
polar  triangle,  that  is,  a  tri- 
angle whose  vertices  are  the 
poles  of  the  sides  of  ABC. 
Prolong  the  sides  AB^  A  C, 
and  BC  till  they  meet  the 
sides  of  DEF.  The  angle 
A  will  be  measured  by  the 
arc  GH  (No.  124,  Cor.  I.) 
Now,  since  EJI=:90°  and 
GF=:GJI+RF=  90°,  we 
have 

EF=ER+IIF=90°+{90°-GR\or  GR=1S0°-EF, 

and  therefore  ^  =  180°  -  EF 


GEOMETRY, 


125 


And  in  a  similar  manner  it  may  be  shown  that 

^  —  180°-  DF,  and  C^  180°-  DK    Q.  E.  D. 

Cor.  From  the  definition 
of  polar  triangles  it  is  ob- 
vious that  if  a  triangle 
A'B'C  is  polar  to  the  tri- 
angle ABG,  then  ABC  is 
also  polar  to  A' B' C .  Hence 
calling  «,  J,  G  and  a' ^  h\  c' 
the  sides  opposite  to  the  an- 
gles ^,  ^,  G2.xi^A',B\C', 
we  shall  liave  not  only 

JL=zl80°-^',      ^  =  180°- 
but  also 
A'=i  180°-  a,      B'=  180°- 1, 


(7'=:180°-C. 


PKOPOSITIOlSr  YI. 

139.  The  sum  of  the  sides  of  a  spherical  triangle  is 
always  less  than  the  circumference  of  a  great  circle. 

Proof  Let  ABC  be  a  spherical  triangle.  The  sides 
AB  and  AC  produced  will 
meet  at  a  point  D  diametri- 
cally opposite  to  A;  hence 
the  two  arcs  ABD  and  A  CD 
together  are  the  circumference 
of  a  great  circle.  But  {^o. 
136)  BC  is  less  than  BD-\- 

CD ;  therefore  the  sum  ^J?-j-^^+^^  is  less  than 
the  circumference  of  a  great  circle.     Q.  E.  D. 


126 


ELEMENTS  OF 


PEOPOSITION  yii. 

140.  Symmetrical  triangles  are  equal  in  all  their 
jparts. 

Proof.  Triangles  are  called  symmetrical  when  they 
have  sides  respectively  equal,  but  disposed  in  opposite 
ways,  so  that  one  triangle  cannot  be  made  to  coincide 
with  the  other  by  superposition. 
Let  ABC  and  BCD  be  two  sym- 
metrical triangles  having  AB^=^ 
BD,  AC—  CD,  and  the  third  side 
BC  oi  the  one  placed  on  the  equal 
side  BC  oi  the  other.  Draw  the 
radii  AG,  BO,  CO,  DO.  The  ra- 
dii AG,  BO,  CO  will  be  the  edges 
of  a  trihedral  angle  having  its  ver- 
tex at  0,  whilst  the  radii  BO,  DO, 
CO  will  be  the  edges  of  a  second 
trihedral  angle  having  also  its  ver- 
tex at  0.  As  the  plane  angles  formed  by  these  edges 
are  respectively  measured  by  equal  arcs,  they  are  respec- 
tively equal ;  hence  (No.  100)  the  planes  of  the  equal 
angles  are  equally  inclined  to  each  other.  But  (No.  134) 
tlie  angles  made  by  these  planes  are  equal  to  the  corre- 
sponding angles  of  the  spherical  triangles.  We  have, 
therefore,  ^^(7=  ^i> 6/,  ABC=DBC,  and  ACB — 
DCB  ;  and  consequently  the  two  triangles  are  equal  in 
all  their  parts.     Q.  E.  D. 

Cor.  If  the  two  triangles  have  angles  respectively 
equal,  their  sides  will  also  be  respectively  equal.  For 
if  the  angles  are  respectively  equal,  the  planes  which 
form  them  are  equally  inclined  to  each  other,  and  their 
edges  form  in  0  equal  trihedral  angles,  having  equal 
faces,   that  is,  composed  of  plane   angles  respectively 


GEOMETRY,  127 


equal.  JS'ow,  these  plane  angles  are  measured  by  the 
opposite  arcs,  that  is,  by  the  sides  of  the  two  triangles. 
And  therefore  the  sides  opposite  to  equal  angles  will  be 
equal,  and  the  two  triangles  will  be  equal  in  all  their 
parts. 


PKOPOsiTiojsr  yiii. 

141.  Two  spherical  triangles  having  two  sides  and 
the  included  angle  of  the  one  equal  to  two  sides  and  the 
included  angle  of  the  other,  each  to  each,  are  equal  in  all 
their  parts. 

Proof.  Let  AB  G  and  DEF  have  the  sides  AB  -DE, 
and  AC^=.DE,  and  the  included  angles  A  and  D  equal. 
Since  A  —  D,  if  the  side 
DE  of  the  second  triangle 
be  superposed  on  the  side 
AG  oi  the  first,  the  side 
DE  will  necessarily  coin- 
cide with  the  side  AB,  and 
the'  point  E  will  fall  on  B  ; 
and  consequently  the  side 
EE  will  lie  on  the  side  BG.  The  two  triangles  are 
therefore  equal  in  all  their  parts.  If  the  triangles  can- 
not be  superposed  owing  to  their  symmetrical  position, 
as  is  the  case  with  the  triangles  ^^6^  and  DEG,  then 
by  constructing  the  triangle  DEE  symmetrical  to  DEG, 
and  by  superposing  it  on  ABG,  we  first  show  the  equali- 
ty of  i>^i^  with  ABG,  then,  because  DEE  is  equal  to 
its  symmetrical  DEG,  we  conclude  that  DFG  also  is 
equal  to  ^^(7.     Q.  E.  D. 


128  ELEMENTS  OF 


PEOPOSITIOlSi    IX. 

142.  Two  spherical  triangles  having  two  angles  and 
the  included  side  of  the  one  equal  to  two  angles  and  the 
included  side  of  the  other ^  each  to  each,  are  equal  in  all 
their  jparts. 

Proof  Proceeding  as  in  the  last  proposition,  place  the 
side  DF Vi^oi\  its  equal  AC.  If  the  angles  A  and D  are 
equal  the  side  DE  will  lie  upon  AB,  and  if  the  angles 
F  and  (7' are  equal  the  side  FE  will  fall  upon  BC, 
and  the  intersection  of  the  arcs  DF  and  FF  will  coin- 
cide with  the  intersection  of  AB  and  BC,  and  therefore 
the  triangles  will  entirely  coincide.  If  the  triangles  can- 
not be  superposed,  as  is  the  case  with  ABC  and  DFG, 
make  the  triangle  DFF  symmetrical  to  DFG ;  super- 
pose DFF  on  ABC,  as  above,  and  from  its  equality  with 
ABC,  it  will  be  proved  thati>i^6^  also  must  be  equal  to 
ABC     Q.  E.D. 


PKOPOSITIOI^  X. 

143.  Two  spherical  triangles^  lohose  sides  are  all  re- 
spectively equal,  are  equal  in  all  their  jparts. 

Proof.  If  the  two  triangles  are  symmetrical,  it  has  al- 
ready been  shown  (No.  140)  that  equal  angles  will  be 
opposite  the  equal  sides.  If  the  two  triangles  are  not 
symmetrical,  they  may  be  superposed  so  as  to  coincide 
throughout ;  for,  as  their  sides  are  equal  to  those  of  the 
symmetrical  triangles,  so  also  their  angles  will  be  equal 
to  those  of  the  symmetrical  triangles  (No.  140);  and 
therefore  they  will  be  equal  in  all  their  parts.     Q.  E.  D. 


GEOMETRY.  129 


PEOPOSITION  XI. 

144.  In  any  isosceles  sjpherical  triangle  the  angles 
opposite  to  the  equal  sides  are  equal  i  and  conversely. 

Proof.  Let  ABC  be  an  isosceles 
triangle  liaving  AB::^AC.  Draw 
an  arc  of  a  great  circle  from  the  ver- 
tex A  to  the  middle  point  D  of  the 
base  BC.  The  triangles  ABD  and 
ACD  will  have  their  sides  respec- 
tively equal ;  hence  the  angles  B  and 
C  opposite  to  the  same  side  AD  will 
be  equal  (No.  143).     Q.  E.  D. 

Conversely,  if  the  angles  B  and  C 
are  equal,  the  opposite  sides  ^6^  and  AB  must  be  equal. 
For,  if  AB^  for  instance,  were  greater  than  A  (7,  it  might 
be  shortened  by  a  portion^^',  so  as  to  leave  BE^^AG ; 
but  in  such  a  case,  drawing  the  arc  EG^  we  would  have 
in  the  triangles  ABG  and  EBG  the  sides  AG  and  BE 
equal  by  construction,  the  side  ^6^  common,  and  the  in- 
cluded angle  ^6/5  equal  to  the  included  angle  ^^5  (7. 
Hence  (Xo.  141)  the  triangles  J.^(7and  EBG  would  be 
equal  in  all  their  parts,  and  therefore  the  angle  A  GB 
would  be  equal  to  the  angle  EGB,  that  is,  the  whole 
would  be  equal  to  one  of  its  parts.  This  being  impos- 
sible, it  is  impossible  to  admit  that  one  of  the  >  two  sides 
is  greater  than  the  other.     Q.  E.  D. 


PKOPOSITIO:^    XII. 

145.  In  any  spherical  triangle  the  greater  side  is 
opposite  to  the  greater  angle  ;  and  conversely. 

Proof.  Let  ^^{7  be  a  spherical  triangle  in  which  the 


130  ELEMENTS  OF 


angle  A  is  greater  than  the  angle  B.     Draw  the  arc  AD 

so  as  to  make  the  angle 

DAB  equal  to  the  angle 

B.    Then  ABD  will  be 

isosceles,  and  AD  =  BD. 

N^ow,  the  sum  AD  +  CD 

is  greater  than  A  C  (No. 

136).     Therefore  the  sum 

BD  -f-  CD^   or    the    side  BG,  is  greater    than   AG. 

Q.  E.  D. 

Conversely,  if  j56^>  J.(7,  the  angle  A  is  greater  than 
the  angle  B.  For,  if  ^  were  not  greater  than  B,  it 
would  be  either  less  or  equal.  Now,  it  cannot  be  equal, 
so  long  as  the  opposite  sides  are  unequal  (No.  IM) ;  and 
it  cannot  be  less,  for  then  BG  should  be  less  than  AG, 
as  we  have  just  proved. 


PEOPOSITION  XIII. 

146.  Two  spherical  triangles  having  all  their  angles 
respectively  equal,  are  equal  in  all  their  parts. 

Proof.  If  the  two  triangles  are  symmetrical,  it  has  al- 
ready been  shown  (No.  140,  Cor.)  that  equal  sides  are 
opposite  their  equal  angles.  If  the  two  triangles  are  not 
symmetrical,  they  may  be  superposed  so  as  to  coincide 
throughout;  for,  as  their  angles  are  equal  to  those  of 
the  symmetrical  triangles,  so  also  their  sides  will  be 
equal  to  those  of  the  symmetrical  triangles  (No.  140, 
Cor.) ;  and  therefore  they  will  be  equal  in  all  their 
parts.     Q.  E.  D. 

Gor.  Two  spherical  triangles  cannot  he  similar  with- 
out being  equal. 


GEOMETRY. 


131 


PKOPOSITIO]^  XIY. 

147.  The  sum  of  the  angles  of  a  sjpherical  triangle 
is  greater  than  two  right  angles,  and  less  than  six  right 
angles. 

Proof.  JuQtABC2.ndLDEF 
be  two  polar  triangles.  We 
liave  seen  that  the  angle  A 
is  measured  by  180°—  EF,  the 
angle  B  by  180°-  DF,  the  an- 
gle C  by  1S0°-I)F.  Hence 
each  angle  is  less  than  180°; 
and  therefore  their  sum  is  less 
than  3x180°,  or  six  right  an- 
gles.    On  the  other  hand,  since 

A  +  B+C=SxlSO''-{FF+J)F+I)F), 

and  the  sum  EF-\-  DF-{-  DE  is  necessarily  less  than 
360°  (No.  139),  it  follows  that  A-^B-^G  is  greater 
than  3x180°— 360°,  that  is,  is  greater  than  180°,  or  two 
right  angles.     Q.  E.  D. 

Cor.  I.  A  spherical  triangle  may  have  two,  and  even 
three,  right  angles;  also  two,  and  even  three,  obtuse 
angles. 

Cor.  II.  If  a  spherical  triangle  is  tri-rectangular,  its 
sides  will  all  be  quadrants  of  a  great  circle.  For  if  in  a 
triangle  AB  C  we  have  A  =  90°,  B  =  90°,  C=z  90°,  then 
in  its  polar  triangle  we  must  have  (No.  138,  Cor.) 

«'=180°-^=:90°, 

5'=  180°- ^=90°, 

c'— 180°-  ^=:90°, 


and  thus,  the  sides  a\  V,  c'  being  quadrants,  the  angles 


132  ELEMENTS  OF 


A',  B\  C  of  the  triangle  A'B'C  will  be  tlie  poles  of 
this  triangle  as  well  as  of  the  triangle  ABC.  Ilence  the 
two  polar  triangles  are,  in  this  case,  one  and  the  same 
triangle,  and  a^a\  h  =  h\  c  =  c^  /  that  is,  all  the  sides 
are  quadrants.  Conversely,  if  all  sides  be  quadrants,  the 
triangle  will  be  tri-rectangular. 

Cor.  III.  Four  tri-rectangular  triangles  make  up  the 
surface  of  a  hemisphere,  and  eight  the  entire  surface  of 
the  sphere.  Hence  the  area  of  a  tri-rectangular  triangle 
is  expressed  (No.  126,  Cor.  I.)  by  T^  ^Tzr". 

Cor.  lY.  If  a  spherical  triangle  is  bi-rectangular,  the 
sides  opposite  to  the  right  angles  will  be  quadrants  of  a 
circle ;  for,  since  they  are  perpendicular  to  the  third 
side,  they  must  pass  through 
the  pole  of  that  side  (No.  135, 
Cor.  lY.),  and  their  intersection 
will  take  place  at  the  pole  it- 
self. Thus  if  ABC  is  bi-rectan- 
gular, the  sides  AB  and  A  (7,  op- 
posite to  the  right  angles,  are 
quadrants,  and  A  is  the  pole  of 
the  side  BC.  Conversely,  if  the 
sides  AB  and  AC  are  quadrants,  the  angles  C  and  B 
will  be  right  angles;  for  the  planes  AOB  and  AOG 
will  be  perpendicular  to  the  plane  BOC 


PEOPOSITION    XY. 

148.  The  surface  of  a  lune  is  to  the  surface  of  the 
sphere  as  its  angle  is  to  four  right  angles. 

Proof,  Conceive  the  surface  of  the  sphere  as  made  up 


GEOMETRY. 


133 


of  360  lunes  having  each  an  angle  of  1°.    Then  360  such 
lunes  will  be  equal  to  the  surface  of  the  sphere.     Hence 
designating  one  of  these  lunes 
by  Z,  and  the  surface  of  the 
sphere  by  S^  we  shall  have 

or        Z:a^::  1:360. 

Let  us  have  now  any  lune 
ABECA^  and  suppose  it  to 
contain  m  times  the  lune  I; 
m  being  any  entire  or  any 
fractional   number  whatever. 

Its  angle  will  also  be  m  times  the  angle  1°,  that  is, 
m  degrees.  The  preceding  proportion  will  then  be- 
come 

that  is,  the  surface  of  the  lune  is  to  that  of  the  sphere  as 
the  angle  of  the  lune  is  to  four  right  angles.     Q.  E.  D. 

Cor,  If  we  denote  the  area  of  a  tri-rectangular  triangle 
by  T,  the  area  of  the  lune  by  Z,  and  the  angle  of  the 
lune  by  J.,  the  right  angle  being  denoted  by  1,  we  shall 
have 

Z:8r::J.:4: 


hence 


L=:2AxT; 


that  is,  the  surface  of  a  lune  is  equal  to  a  tri-rectangular 
triangle  multiplied  by  twice  the  angle  of  the  lune.  As 
90°  is  here  made  =1,  the  value  of  the  angle  A  must 
be  the  expression  of  how  many  times  A  degrees  con- 
tain 90°. 


134 


ELEMENTS  OF 


PKOPOSITIOJSr  XYI. 

14:9,  The  area  of  a  spherical  triangle  is  equal  to  its 
spherical  excess  multiplied  hy  a  tri-rectangular  triangle. 

Proof.  By  spherical  excess  we  mean  the  excess  of  the 
sum  of  the  angles  of  a  spherical  triangle  over  two  right 
angles.  This  excess  is  calculated  by  taking  the  right  angle 
as  the  unit  of  measure  of  spherical  angles ;  and  consequent- 
ly the  number  of  degrees  contained  in  the  spherical  ex- 
cess must  be  divided  by  90°  ;  and  the  quotient,  which  is 
the  true  value  of  the  spherical  excess,  is  to  be  considered 
an  abstract  number  or  coefficient. 

Let,  then,  ABC  be  any  spherical  triangle  drawn  on 
the  hemisphere  DEHFGK.  Prolong  the  arcs  AB^  A  (7, 
BG  till  they  reach  the 
section  of  the  hemi- 
sphere. Then  the  ver- 
tical triangles  ABE 
and  AGF  taken  to- 
gether will  be  equal  to 
a  lune  of  the  angle  A^ 
the  triangles  BFII  and 
BBK  taken  together 
will  be  equal  to  a  lune 
of  the  angle  B^  and  the 
triangles  GGK  and 
GEH  taken  together 
will  be  equal  to  a  lune 
of  the  angle  G.  The  sum  of  the 
(Ko.  148,  Cor.) 


three  lunes  will  be 


2^xr+2^X^+2(7xr=2r(^  +  ^-f  ^). 


Now,  this  sum  exceeds  the  surface  4T  of  the  hemisphere 


GEOMETRY.  135 


by  twice  the  triangle  ABC ;  for  this  triangle  has  been 
taken  three  times  instead  of  one  time.  We  have  there- 
fore 

2ABC=2T{A  +  B-\-C)-4.T. 

Dividing  the  equation  by  2,  and  reducing,  we  have 

ABC=:{A-{-B+C-2)T, 

A -\- B  -\- C  — '^  being  the  spherical  excess.     Q.  E.  D. 

Cor.  Since  the  area  of  the  triangle  has  been  deduced 
from  the  area  of  the  lunes,  in  whose  expression  an  angle 
of  90°  was  taken  as  a  unit,  the  number  2  in  the  expres- 
sion of  the  spherical  excess  represents  180  degrees.  Thus, 
if  in  a  given  triangle  we  have  A  —  86°  10^,  B  —  69°  20', 
^=100°  15',  the  value  of  the  spherical  excess  will  be 
expressed  by 

86°  10'+  69°  20'+ 100°  15'-  180° 
90°  ' 

the  quotient  0.841666  .  .  .  being  an  abstract  number. 


PROPOSITIOlSr  XYII. 

150,  The  area  of  a  spherical  polygon  is  equal  to  its 
spherical  excess  multiplied  hy  a  tri-rectangular  triangle. 

Proof.  A  spherical  polygon 
ABCDE  can  always  be  divided 
into  triangles  by  drawing  diago- 
nal arcs.  If  the  polygon  has  n 
sides,  it  will  be  divided  into  n— 2 
triangles,  every  one  of  which  will 
be  equal  to  its  spherical  excess 
multiplied    by    a    tri-rectangular 


136 


ELEMENTS  OF 


triangle  (IS'o.  149).     Hence  tlieir  sum,  or  the  area  of  the 
polygon,  will  be  equal  to 

S  being  the  sum  of  all  the  angles.     Q.  E.  D. 


PROPOSITION  XYIII. 

151.  The  volume  of  a  sj)herieal  wedge  is  equal  to  the 
product  of  its  base  [the  Iwie)  into  one-third  of  the 
radius. 

Proof.  The  volume  of  the  sphere  being  conceived  as 
made  up  of  spherical  wedges,  just  as  its 
surface  is  conceived  as  made  up  of  as 
many  lunes,  we  easily  perceive  that  the 
wedge  is  to  the  volume  of  the  sphere  as 
its  base,  or  lune,  is  to  the  surface  of  tlie 
sphere.     Hence,  W  being  the  wedge, 

AiKr^ 


W  \ 


\L\  ^7:r'; 


which  is  readily  reduced  to 

Wz 


Z  .  ^.    Q.  E.  D. 

o 


PROPOSITION    XIX. 

152.  The  volume  of  a  spherical  pyramid  is  equal  to 
the  ^product  of  its  hase  {a  spherical  polygon]  into  one- 
third  of  the  radius. 

Proof  Let  ABCD  —  0  be  a  spherical  pyramid.    Di- 


GEOMETRY. 


137 


viding  the  area  of  the  base  ABCD  into  a  very  great 
number  of  exceedingly  small  areas,  the  spherical  pyra- 
mid may  be  conceived  as  a  sum  of  pyramids  of  an  alti- 
tude equal  to  the  radius  of  the  sphere,  and  having  for 
their  bases  the  small  areas  of  which 
the  polygon  ABCD  is  composed. 
These  areas,  when  exceedingly  small, 
do  not  differ  from  plane  surfaces ; 
and  therefore  (No.  113,  Cor.  II.)  the 
volume  of  each  small  pyramid  will 
be  equal  to  the  product  of  its  small 
base  into  one-third  of  the  radius. 
Consequently  the  spherical  pyramid, 
which  is  the  sum  of  all  the  small 
pyramids,  will  be  equal  to  the  sum 
of  all  the  small  bases  multiplied  by 
one-third  of  the  radius.  But  the  sum 
of  all  the  small  bases  equals  the  base  ABCD.  There- 
fore the  volume  of  the  spherical  pyramid  is  equal  to  the 
product  of  its  base  into  one-third  of  the  radius  of  the 
sphere.     Q.  E.  D. 


14  JJAY    Ubh 

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